Chapter 7 Mean Value Theorems-I

In the theorems that follow in this chapter, it is essential to know if f(x)$f\left(x\right)$ is continuous or not, derivable or not in the given interval.

7.0.0.1 Which functions can we expect to be continuous and differentiable?

1. Every algebraic polynomial in x$x$ is a continuous function of x$x$ for every value of x$x$.
2. sinx$\sin x$, cosx$\cos x$, ex$e^x$ are continuous for all values of x$x$ and logx$\log x$ is continuous ∀$\forall$ x>0$x>0$.
3. If f$f$ and g$g$ are two functions in closed interval [a,b]$[a,b]$, then f±g$f\pm g$, f×g$f\times g$ are also continuous in [a,b]$[a,b]$ and fg$\frac{f}{g}$ is also continuous in that interval provided g(x)≠0$g\left(x\right)\ne 0$ for any x∈[a,b]$x\in [a,b]$.
4. If a function is differentiable for every point in a given interval, then it must be continuous in that interval.
5. Differentiability implies continuity but reverse is not true.

7.0.0.2 How to check for continuity of function?

• A function f(x)$f\left(x\right)$ is continuous at point x=a$x=a$ if limx→a−f(x)=limx→a+f(x)=f(a)${lim}_{x\to a^{-}}f\left(x\right)={lim}_{x\to a^{+}}f\left(x\right)=f\left(a\right)$. We know that limx→a−f(x)=limh→0f(a−h)${lim}_{x\to a^{-}}f\left(x\right)={lim}_{h\to 0}f(a-h)$ and limx→a+f(x)=limh→0f(a+h)${lim}_{x\to a^{+}}f\left(x\right)={lim}_{h\to 0}f(a+h)$.

• A function f(x)$f\left(x\right)$ is discontinuous at point x=a$x=a$ if:

  • limx→af(x)${lim}_{x\to a}f\left(x\right)$ does not exist, or
  • limx→a−f(x)≠limx→a+f(x)${lim}_{x\to a^{-}}f\left(x\right)\ne {lim}_{x\to a^{+}}f\left(x\right)$, or
  • limx→a−f(x)=limx→a+≠f(a)${lim}_{x\to a^{-}}f\left(x\right)={lim}_{x\to a^{+}}\ne f\left(a\right)$, or
  • f(x)$f\left(x\right)$ does not exist at x=a$x=a$

7.0.0.3 How to check for differentiabity of function?

Calculate the derivative of the given function in the given open interval (a,b)$(a,b)$. If it is not

• infinite
• imaginary, or
• indeterminate,

then its derivable.

Differentiability implies continuity but reverse is not true.

Rolle’s theorem

If a function f(x)$f\left(x\right)$ is

1. continuous in the closed interval [a,b]$[a,b]$
2. derivable in the open interval (a,b)$(a,b)$
3. f(a)=f(b)$f\left(a\right)=f\left(b\right)$

then there exists at least one value c∈(a,b)$c\in (a,b)$ such that f′(c)=0$f^{\prime }\left(c\right)=0$.

7.1 Exercise 4 (i)

7.1.1 Question 1

Verify Rolle’s theorem for:

• f(x)=x2,x∈[−1,1]$f\left(x\right)=x^2,x\in [-1,1]$

The given function f(x)$f\left(x\right)$ is a polynomial function, so it is continuous in [−1,1]$[-1,1]$ and derivable ∀$\forall$ values of x∈(−1,1)$x\in (-1,1)$.

f(−1)=1=f(1)$$\begin{array}{cc}f(-1)& =1=f\left(1\right)\end{array}$$

f′(x)=2x$$\begin{array}{cc}{f}^{\prime }\left(x\right)& =2x\end{array}$$

So as the three conditions of Rolle’s theorem is satisfied by the equation, then there must exist atleast a point c∈(−1,1)$c\in (-1,1)$ such that f′(c)=0$f^{\prime }\left(c\right)=0$.

2c=0c=0$$\begin{array}{cc}2c& =0\\ c& =0\end{array}$$

c=0$c=0$ lies in given interval (−1,1)$(-1,1)$. Thus Rolle’s theorem is verified.

• f(x)=x(x+3)e−x/2,x∈[−3,0]$f\left(x\right)=x(x+3)e^{-x/2},x\in [-3,0]$

Being the product of a polynomial and exponential function, f(x)$f\left(x\right)$ is continuous in [−3,0]$[-3,0]$.

Also,

f(x)=x(x+3)e−x/2=(x2+3x)e−x/2f′(x)=(2x+3)e−x/2+(x2+3x)e−x/2×−12=e−x/2[2x+3−x2+3x2]f′(x)=−12e−x/2(x2−x−6)$$\begin{array}{cc}f\left(x\right)& =x(x+3)e^{-x/2}\\ & =(x^2+3x)e^{-x/2}\\ f^{\prime }\left(x\right)& =(2x+3)e^{-x/2}+(x^2+3x)e^{-x/2}\times -\frac{1}{2}\\ & =e^{-x/2}[2x+3-\frac{x^2+3x}{2}]\\ f^{\prime }\left(x\right)& =-\frac{1}{2}{e}^{-x/2}(x^2-x-6)\end{array}$$

which is finite ∀$\forall$ x∈(−3,0)$x\in (-3,0)$, hence the given function is derivable in (−3,0)$(-3,0)$.

In the given equation, f(−3)=0=f(0)$$\begin{array}{c}f(-3)=0=f\left(0\right)\end{array}$$

So as the three conditions of Rolle’s theorem is satisfied by the equation, then there must exist atleast a point c∈(−3,0)$c\in (-3,0)$ such that f′(c)=0$f^{\prime }\left(c\right)=0$.

−12e−c/2(c2−c−6)=0c2−c−6=0c=−2,3$$\begin{array}{cc}-\frac{1}{2}{e}^{-c/2}(c^2-c-6)& =0\\ c^2-c-6& =0\\ c& =-2,3\end{array}$$

Only c=−2$c=-2$ lies in the interval (−3,0)$(-3,0)$. Thus one value of c$c$ has been found at which f′(c)=0$f^{\prime }\left(c\right)=0$. Hence, Rolle’s theorem is verified.

• ϕ(x)=log{x2+ab(a+b)x},x∈[a,b]$\varphi \left(x\right)=\log \left\{\frac{x^2+ab}{(a+b)x}\right\},x\in [a,b]$

This is a logarithmic function and is continuous in [a,b]$[a,b]$.

Also,

ϕ(x)=log{x2+ab(a+b)x}=log(xa+b+1xab(a+b))ϕ′(x)=1{x2+ab(a+b)x}(1a+b−ab(a+b)×1x2)ϕ′(x)=(a+b)xx2+ab(x2−abx2(a+b))ϕ′(x)=x2−abx(x2+ab)$$\begin{array}{cc}\varphi \left(x\right)& =\log \left\{\frac{x^2+ab}{(a+b)x}\right\}=\log (\frac{x}{a+b}+\frac{1}{x}\frac{ab}{(a+b)})\\ {\varphi }^{\prime }\left(x\right)& =\frac{1}{\left\{\frac{x^2+ab}{(a+b)x}\right\}}(\frac{1}{a+b}-\frac{ab}{(a+b)}\times \frac{1}{x^2})\\ {\varphi }^{\prime }\left(x\right)& =\frac{(a+b)x}{x^2+ab}\left(\frac{x^2-ab}{x^2(a+b)}\right)\\ {\varphi }^{\prime }\left(x\right)& =\frac{x^2-ab}{x(x^2+ab)}\end{array}$$

which is finite ∀$\forall$ x∈(a,b)$x\in (a,b)$, hence the given function is differentiable in (a,b)$(a,b)$.

ϕ(a)=log{a2+aba(a+b)}=log1=0ϕ(b)=log{b2+abb(a+b)}=log1=0ϕ(a)=0=ϕ(b)$$\begin{array}{cc}\varphi \left(a\right)& =\log \left\{\frac{a^2+ab}{a(a+b)}\right\}=\log 1=0\\ \varphi \left(b\right)& =\log \left\{\frac{b^2+ab}{b(a+b)}\right\}=\log 1=0\\ \varphi \left(a\right)& =0=\varphi \left(b\right)\end{array}$$

So as the three conditions of Rolle’s theorem is satisfied by the equation, then there must exist atleast a point c∈(a,b)$c\in (a,b)$ such that ϕ′(c)=0${\varphi }^{\prime }\left(c\right)=0$.

c2−abc(c2+ab)=0c2=abc=±√ab$$\begin{array}{cc}\frac{c^2-ab}{c(c^2+ab)}& =0\\ c^2& =ab\\ c& =\pm \sqrt{ab}\end{array}$$

Out of two c$c$ values, c=√ab$c=\sqrt{ab}$ lies in the interval (a,b)$(a,b)$ as √ab$\sqrt{ab}$ is the geometric mean of a$a$ and b$b$. Thus Rolle’s theorem is verified.

• ψ(x)=(x−a)m(x−b)n$\psi \left(x\right)=(x-a{)}^m(x-b{)}^n$; m$m$ and n$n$ being positive integers and x∈[a,b]$x\in [a,b]$ (TU 2060)

As m$m$ and n$n$ are positive integers, (x−a)m$(x-a{)}^m$ and (x−b)n$(x-b{)}^n$ are polynomials, thus product of two polynomials is also a polynomial of degree (m+n)$(m+n)$.

Since, every algebraic polynomial in x$x$ is a continuous function of x$x$ for every value of x$x$, f(x)$f\left(x\right)$ is continuous in the interval [a,b]$[a,b]$.

ψ′(x)=(x−a)mn(x−b)n−1+(x−b)nm(x−a)m−1=(x−a)m−1(x−b)n−1[m(x−b)+n(x−a)]$$\begin{array}{cc}{\psi }^{\prime }\left(x\right)& =(x-a{)}^{m}n(x-b{)}^{n-1}+(x-b{)}^{n}m(x-a{)}^{m-1}\\ & =(x-a{)}^{m-1}(x-b{)}^{n-1}\left[m\right(x-b)+n(x-a\left)\right]\end{array}$$ The derivative exists in the interval (a,b)$(a,b)$. Hence the given function is differentiable.

Also,

f(a)=0=f(b)$$\begin{array}{cc}f\left(a\right)& =0=f\left(b\right)\end{array}$$

The given function, thus, satisifies all conditions of Rolle’s theorem. There must exist a value c$c$ such that f′(c)=0$f^{\prime }\left(c\right)=0$.

(c−a)m−1(c−b)n−1[m(c−b)+n(c−a)]=0$$\begin{array}{cc}(c-a{)}^{m-1}(c-b{)}^{n-1}\left[m\right(c-b)+n(c-a\left)\right]& =0\end{array}$$

m(c−b)+n(c−a)=0(m+n)c=mb+nac=mb+nam+n$$\begin{array}{cc}m(c-b)+n(c-a)& =0\\ (m+n)c& =mb+na\\ c& =\frac{mb+na}{m+n}\end{array}$$

which is a point within the interval (a,b)$(a,b)$ dividing it in the ratio m:n$m:n$ internally.

Hence Rolle’s theorem is verified.

• f(x)=ex(sinx−cosx)$f\left(x\right)=e^x(\sin x-\cos x)$ in [π/4,5π/4]$[\pi /4,5\pi /4]$

The function ex$e^x$ is continuous as explained above.

The difference of two functions sinx−cosx$\sin x-\cos x$ is also continuous in given domain [π/4,5π/4]$[\pi /4,5\pi /4]$. See figure below:

Figure 7.1: Unit circle, for ordered pair (x,y): cosθ=x$\cos \theta =x$, sinθ=y$\sin \theta =y$

Thus f(x)$f\left(x\right)$ is continuous in [π/4,5π/4]$[\pi /4,5\pi /4]$.

The derivative of the function is,

f′(x)=ex(cosx+sinx)+(sinx−cosx)ex=2exsinx$$\begin{array}{cc}{f}^{\prime }\left(x\right)& =e^x(\cos x+\sin x)+(\sin x-\cos x)e^x\\ & =2e^x\sin x\end{array}$$

which has finite values in domain (π/4,5π/4)$(\pi /4,5\pi /4)$. Hence the function is differentiable.

Also,

f(π/4)=ex(sinπ/4−cosπ/4)=0f(5π/4)=ex(sin5π/4−cos5π/4)=0f(π/4)=0=f(5π/4)$$\begin{array}{cc}f\left(\pi /4\right)& =e^x(\sin \pi /4-\cos \pi /4)=0\\ f\left(5\pi /4\right)& =e^x(\sin 5\pi /4-\cos 5\pi /4)=0\\ f\left(\pi /4\right)& =0=f\left(5\pi /4\right)\end{array}$$

All conditions of Rolle’s theorem are satisfied. There must exist a value c$c$ such that f′(c)=0$f^{\prime }\left(c\right)=0$.

2ecsinc=0ecsinc=0$$\begin{array}{cc}2e^c\sin c& =0\\ e^c\sin c& =0\end{array}$$

For ec=0$e^c=0$, c$c$ is not defined. So,

sinc=0sinc=sinπc=π$$\begin{array}{cc}\sin c& =0\\ \sin c& =\sin \pi \\ c& =\pi \end{array}$$

The c=π$c=\pi$ lies in (π/4,5π/4)$(\pi /4,5\pi /4)$. Hence Rolle’s theorem is verified.

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Lagrange’s Mean Value Theorem

If a function f(x)$f\left(x\right)$ is

1. continuous in the closed interval [a,b]$[a,b]$
2. derivable in the open interval (a,b)$(a,b)$

then there exists at least one value c∈(a,b)$c\in (a,b)$ such that f(b)−f(a)=(b−a)f′(c)$f\left(b\right)-f\left(a\right)=(b-a)f^{\prime }\left(c\right)$.

7.1.2 Question 2

Verify Lagrange’s mean value theorem for:

• f(x)=x3−4x$f\left(x\right)=x^3-4x$ in [−2,2]$[-2,2]$ (TU 2054)

The given function is a polynomial function, so it is continuous in [−2,2]$[-2,2]$ and derivable in (−2,2)$(-2,2)$.

The derivative of the function is,

f′(x)=3x2−4$$\begin{array}{cc}{f}^{\prime }\left(x\right)& =3x^2-4\end{array}$$

By Lagrange’s MVT,

f′(c)=f(b)−f(a)b−a3c2−4=f(2)−f(−2)2+23c2−4=0c=±2√3$$\begin{array}{cc}{f}^{\prime }\left(c\right)& =\frac{f\left(b\right)-f\left(a\right)}{b-a}\\ 3c^2-4& =\frac{f\left(2\right)-f(-2)}{2+2}\\ 3c^2-4& =0\\ c& =\pm \frac{2}{\sqrt{3}}\end{array}$$

Both values of c$c$ lies in (−2,2)$(-2,2)$. Hence, Lagrange’s MVT is verified.

• f(x)=x(x−1)(x−2),x∈[0,1/2]$f\left(x\right)=x(x-1)(x-2),x\in [0,1/2]$ (TU 2058, 2065)

f(x)=x(x2−2x−x+2)=x3−3x2+2x$$\begin{array}{cc}f\left(x\right)& =x(x^2-2x-x+2)\\ & =x^3-3x^2+2x\end{array}$$

The given function is a polynomial function of degree 3$3$, so it is continuous in [0,1/2]$[0,1/2]$ and derivable in (0,1/2)$(0,1/2)$.

The derivative of function is,

f′(x)=3x2−6x+2$$\begin{array}{cc}{f}^{\prime }\left(x\right)& =3x^2-6x+2\end{array}$$

By Lagrange’s MVT,

f′(c)=f(b)−f(a)b−a3c2−6c+2=f(12)−f(0)12−03c2−6c+2=1/8−3/4+2/21/2=3412c2−24c+5=0$$\begin{array}{cc}{f}^{\prime }\left(c\right)& =\frac{f\left(b\right)-f\left(a\right)}{b-a}\\ 3c^2-6c+2& =\frac{f\left(\frac{1}{2}\right)-f\left(0\right)}{\frac{1}{2}-0}\\ 3c^2-6c+2& =\frac{1/8-3/4+2/2}{1/2}=\frac{3}{4}\\ 12c^2-24c+5& =0\end{array}$$

Finding values of c$c$,

c=−b±√b2−4ac2a=24±√(−24)2−4×12×52×12=24±√33624=24±4√2124=1±√216$$\begin{array}{cc}c& =\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ & =\frac{24\pm \sqrt{(-24{)}^2-4\times 12\times 5}}{2\times 12}\\ & =\frac{24\pm \sqrt{336}}{24}\\ & =\frac{24\pm 4\sqrt{21}}{24}\\ & =1\pm \frac{\sqrt{21}}{6}\end{array}$$

The value of c$c$ that lies in (0,1/2)$(0,1/2)$ is 1−√216$1-\frac{\sqrt{21}}{6}$ which is ≈0.24$\approx 0.24$.

Hence, Lagrange’s MVT is verified.

• f(x)=Ax2+Bx+C,x∈[a,b]$f\left(x\right)=A{x}^2+Bx+C,x\in [a,b]$ (TU 2064)

The given function is a polynomial function of degree 2$2$, so it is continuous in [a,b]$[a,b]$ and derivable in (a,b)$(a,b)$.

The derivative of function is,

f′(x)=2Ax+B$$\begin{array}{cc}{f}^{\prime }\left(x\right)& =2Ax+B\end{array}$$

By Lagrange’s MVT,

f′(c)=f(b)−f(a)b−a=Ab2+Bb+C−Aa2−Ba−Cb−a2Ac+B=A(b2−a2)+B(b−a)b−a2Ac+B=(b−a)(Ab+Aa+B)b−a2Ac+B=Ab+Aa+Bc=a+b2$$\begin{array}{cc}{f}^{\prime }\left(c\right)& =\frac{f\left(b\right)-f\left(a\right)}{b-a}\\ & =\frac{A{b}^2+Bb+C-A{a}^2-Ba-C}{b-a}\\ 2Ac+B& =\frac{A(b^2-a^2)+B(b-a)}{b-a}\\ 2Ac+B& =\frac{(b-a)(Ab+Aa+B)}{b-a}\\ 2Ac+B& =Ab+Aa+B\\ c& =\frac{a+b}{2}\end{array}$$

which is arithmetic mean of a$a$ and b$b$ and lies in open interval (a,b)$(a,b)$. Hence Lagrange’s MVT is verified.

• f(x)=logx$f\left(x\right)=\log x$ in [1,e]$[1,e]$

The given function is logarithmic function, so is continuous in [1,e]$[1,e]$ and derivable in (1,e)$(1,e)$. By Lagrange’s MVT,

f′(c)=f(b)−f(a)b−a1c=loge−log1e−1c=e−1$$\begin{array}{cc}{f}^{\prime }\left(c\right)& =\frac{f\left(b\right)-f\left(a\right)}{b-a}\\ \frac{1}{c}& =\frac{\log e-\log 1}{e-1}\\ c& =e-1\end{array}$$

which lies in interval (1,e)$(1,e)$. Hence Lagrange’s MVT is verified.

• f(x)=√4−x2$f\left(x\right)=\sqrt{4-x^2}$ in [−2,1]$[-2,1]$

The derivative of the function is,

f′(x)=12√4−x2×(−2x)f′(x)=−x√4−x2$$\begin{array}{cc}{f}^{\prime }\left(x\right)& =\frac{1}{2\sqrt{4-x^2}}\times (-2x)\\ f^{\prime }\left(x\right)& =-\frac{x}{\sqrt{4-x^2}}\end{array}$$

The derivative has finite values in interval (−2,1)$(-2,1)$. So the function is derivable in (−2,1)$(-2,1)$. As differentiability implies continuity, the given function is also continuous in (−2,1)$(-2,1)$.

Now let us test the function for continuity at x=−2$x=-2$ and x=1$x=1$.

At x=−2$x=-2$,

limx→(−2)−√4−x2=0=f(−2)$$\begin{array}{cc}\underset{x\to (-2{)}^{-}}{lim}\sqrt{4-x^2}& =0=f(-2)\end{array}$$

At x=1$x=1$,

limx→1+√4−x2=√3=f(1)$$\begin{array}{cc}\underset{x\to 1^{+}}{lim}\sqrt{4-x^2}& =\sqrt{3}=f\left(1\right)\end{array}$$

The right hand limit at x=−2$x=-2$ and left hand limit at x=1$x=1$ is not necessary here, because the function is derivable in (−2,1)$(-2,1)$ and thus those limits are defined.

Thus, the given function is continuous in closed interval [−2,1]$[-2,1]$. By Lagrange’s MVT, there must be a value c$c$ such that,

f′(c)=f(b)−f(a)b−a$$\begin{array}{cc}{f}^{\prime }\left(c\right)& =\frac{f\left(b\right)-f\left(a\right)}{b-a}\end{array}$$

−c√4−c2=√4−12−√4−(−2)21+2−c√4−c2=√3−03$$\begin{array}{cc}-\frac{c}{\sqrt{4-c^2}}& =\frac{\sqrt{4-1^2}-\sqrt{4-(-2{)}^2}}{1+2}\\ -\frac{c}{\sqrt{4-c^2}}& =\frac{\sqrt{3}-0}{3}\end{array}$$

Squaring both sides,

c24−c2=393c2=4−c2c2=1c=±1$$\begin{array}{cc}\frac{c^2}{4-c^2}& =\frac{3}{9}\\ 3c^2& =4-c^2\\ c^2& =1\\ c& =\pm 1\end{array}$$

Thus value of c=−1$c=-1$, lies in (−2,1)$(-2,1)$. Hence, Lagrange’s MVT is verified.

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Cauchy’s Mean Value Theorem

If f(x)$f\left(x\right)$ and F(x)$F\left(x\right)$ are two functions such that,

1. f(x)$f\left(x\right)$ and F(x)$F\left(x\right)$ are continuous in the closed interval [a,b]$[a,b]$
2. f(x)$f\left(x\right)$ and F(x)$F\left(x\right)$ are derivable in the open interval (a,b)$(a,b)$
3. F′(x)≠0$F^{\prime }\left(x\right)\ne 0$ for all (∀$\forall$) x∈(a,b)$x\in (a,b)$

then there exists at least one value c∈(a,b)$c\in (a,b)$ such that

f(b)−f(a)F(b)−F(a)=f′(c)F′(c)$\frac{f\left(b\right)-f\left(a\right)}{F\left(b\right)-F\left(a\right)}=\frac{f^{\prime }\left(c\right)}{F^{\prime }\left(c\right)}$.

7.1.3 Question 3

Verify Cauchy’s Mean Value Theorem for

• f(x)=x2$f\left(x\right)=x^2$ and F(x)=x4$F\left(x\right)=x^4$ in [−1,3]$[-1,3]$

The two given functions are polynomials and hence continuous in [−1,3]$[-1,3]$ and derivable in (−1,3)$(-1,3)$, so the conditions of Cauchy’s MVT are satisfied. Using Cauchy’s MVT, we have,

f(b)−f(a)F(b)−F(a)=f′(c)F′(c)9−181−1=2c4c3880=12c2110=12c2c=±√5$$\begin{array}{cc}\frac{f\left(b\right)-f\left(a\right)}{F\left(b\right)-F\left(a\right)}& =\frac{f^{\prime }\left(c\right)}{F^{\prime }\left(c\right)}\\ \frac{9-1}{81-1}& =\frac{2c}{4c^3}\\ \frac{8}{80}& =\frac{1}{2c^2}\\ \frac{1}{10}& =\frac{1}{2c^2}\\ c& =\pm \sqrt{5}\end{array}$$

The value of c=√5$c=\sqrt{5}$ lies in the interval (−1,3)$(-1,3)$. Hence, Cauchy’s MVT is verified.

• f(x)=sinx$f\left(x\right)=\sin x$ and F(x)=cosx$F\left(x\right)=\cos x$ in [0,π/2]$[0,\pi /2]$

The two functions are trigonometric functions and are continuous in [0,π/2]$[0,\pi /2]$. The derivatives f′(x)=cosx$f^{\prime }\left(x\right)=\cos x$ and F′(x)=−sinx$F^{\prime }\left(x\right)=-\sin x$ are also finite for all x∈(0,π/2)$x\in (0,\pi /2)$. Hence the given functions are also derivable.

Figure 7.2: Unit circle, for ordered pair (x,y): cosθ=x$\cos \theta =x$, sinθ=y$\sin \theta =y$

So the conditions of Cauchy’s MVT are satisfied. Using Cauchy’s MVT, we have,

f(b)−f(a)F(b)−F(a)=f′(c)F′(c)sinπ/2−sin0cosπ/2−cos0=−coscsinc1−00−1=−coscsinctanc=1c=π4$$\begin{array}{cc}\frac{f\left(b\right)-f\left(a\right)}{F\left(b\right)-F\left(a\right)}& =\frac{f^{\prime }\left(c\right)}{F^{\prime }\left(c\right)}\\ \frac{\sin \pi /2-\sin 0}{\cos \pi /2-\cos 0}& =\frac{-\cos c}{\sin c}\\ \frac{1-0}{0-1}& =-\frac{\cos c}{\sin c}\\ \tan c& =1\\ c& =\frac{\pi }{4}\end{array}$$

which lies in open interval (0,π/2)$(0,\pi /2)$. This verifies Cauchy’s MVT.

7.1.4 Question 4

1. In the Mean Value Theorem f(x+h)=f(x)+hf′(x+θh)$f(x+h)=f\left(x\right)+h{f}^{\prime }(x+\theta h)$ find θ$\theta$ if f(x)=Ax2+Bx+C,A≠0$f\left(x\right)=A{x}^2+Bx+C,A\ne 0$

f(x)=Ax2+Bx+Cf(x+h)=A(x+h)2+B(x+h)+Cf′(x)=2Ax+Bf′(x+θh)=2A(x+θh)+B$$\begin{array}{ccc}f\left(x\right)& =A{x}^2+Bx+Cf(x+h)& =A(x+h{)}^2+B(x+h)+C\\ f^{\prime }\left(x\right)& =2Ax+B& \\ f^{\prime }(x+\theta h)& =2A(x+\theta h)+B& \end{array}$$

The MVT is f(x+h)=f(x)+hf′(x+θh)$f(x+h)=f\left(x\right)+h{f}^{\prime }(x+\theta h)$, so substituting relevant values,

A(x+h)2+B(x+h)+C=Ax2+Bx+C+h(2A(x+θh)+B)Ax2+2Ahx+Ah2+Bx+Bh+C=Ax2+Bx+C+2Ahx+2Ah2θ+BhAh2=2Ah2θθ=12$$\begin{array}{cc}A(x+h{)}^2+B(x+h)+C& =A{x}^2+Bx+C+h\left(2A\right(x+\theta h)+B)\\ A{x}^2+2Ahx+A{h}^2+Bx+Bh+C& =A{x}^2+Bx+C+2Ahx+2A{h}^2\theta +Bh\\ A{h}^2& =2A{h}^2\theta \\ \theta & =\frac{1}{2}\end{array}$$

Thus, θ=12$\theta =\frac{1}{2}$ which is 0<θ<1$0<\theta <1$.

2. In the Mean Value Theorem f(a+h)=f(a)+hf′(a+θh)$f(a+h)=f\left(a\right)+h{f}^{\prime }(a+\theta h)$ find θ$\theta$ if a=1,h=3,f(x)=√x$a=1,h=3,f\left(x\right)=\sqrt{x}$.

Here,

f(x)=√xf(a)=√af(a+h)=√a+hf′(x)=12√xf′(a+θh)=12√a+θh$$\begin{array}{cc}f\left(x\right)& =\sqrt{x}\\ f\left(a\right)& =\sqrt{a}\\ f(a+h)& =\sqrt{a+h}\\ f^{\prime }\left(x\right)& =\frac{1}{2\sqrt{x}}\\ f^{\prime }(a+\theta h)& =\frac{1}{2\sqrt{a+\theta h}}\end{array}$$

The Mean Value Theorem is f(a+h)=f(a)+hf′(a+θh)$f(a+h)=f\left(a\right)+h{f}^{\prime }(a+\theta h)$, substituting respective values from above,

√a+h=√a+h(12√a+θh)$$\begin{array}{cc}\sqrt{a+h}& =\sqrt{a}+h\left(\frac{1}{2\sqrt{a+\theta h}}\right)\end{array}$$

Substituting a=1,h=3$a=1,h=3$,

√1+3=√1+3(12√1+3θ)√4−√1=32√1+3θ$$\begin{array}{cc}\sqrt{1+3}& =\sqrt{1}+3\left(\frac{1}{2\sqrt{1+3\theta }}\right)\\ \sqrt{4}-\sqrt{1}& =\frac{3}{2\sqrt{1+3\theta }}\end{array}$$

Squaring both sides,

4−2√4+1=94(1+3θ)5±4=94(1+3θ)$$\begin{array}{cc}4-2\sqrt{4}+1& =\frac{9}{4(1+3\theta )}\\ 5\pm 4& =\frac{9}{4(1+3\theta )}\end{array}$$

Taking positive sign,

5+4=94(1+3θ)4+12θ=1θ=−312$$\begin{array}{cc}5+4& =\frac{9}{4(1+3\theta )}\\ 4+12\theta & =1\\ \theta & =-\frac{3}{12}\end{array}$$

Taking negative sign,

5−4=94(1+3θ)4+12θ=9θ=512$$\begin{array}{cc}5-4& =\frac{9}{4(1+3\theta )}\\ 4+12\theta & =9\\ \theta & =\frac{5}{12}\end{array}$$

So we have θ=−312,512$\theta =-\frac{3}{12},\frac{5}{12}$, but only 512$\frac{5}{12}$ lies in (0,1)$(0,1)$ as required by alternate form of Lagrange’s MVT. Hence the required value of θ=512$\theta =\frac{5}{12}$.

3. Find the values of θ$\theta$ in the Mean Value Theorem if

   • f(x)=ex$f\left(x\right)=e^x$

   The function f(x)$f\left(x\right)$ is continuous in [a,a+h]$[a,a+h]$ and differentiable in (a,a+h)$(a,a+h)$. There must exist a value θ$\theta$ such that f(a+h)=f(a)+hf′(a+θh)$f(a+h)=f\left(a\right)+h{f}^{\prime }(a+\theta h)$ where 0<θ<1$0<\theta <1$. This is the alternate form of Lagrange’s MVT.

   f(x)=exf(a+h)=ea+hf(a)=eaf′(x)=exf′(a+θh)=ea+θh$$\begin{array}{cc}f\left(x\right)& =e^x\\ f(a+h)& =e^{a+h}\\ f\left(a\right)& =e^a\\ f^{\prime }\left(x\right)& =e^x\\ f^{\prime }(a+\theta h)& =e^{a+\theta h}\end{array}$$

   Substituting the values in alternate form of Lagrange’s MVT, we have,

   ea+h=ea+hea+θheh=1+heθheθh=eh−1h$$\begin{array}{cc}{e}^{a+h}& =e^a+h{e}^{a+\theta h}\\ e^h& =1+h{e}^{\theta h}\\ e^{\theta h}& =\frac{e^h-1}{h}\end{array}$$

   Taking log both sides,

   θh=log(eh−1h)θ=1hlog(eh−1h)$$\begin{array}{cc}\theta h& =\log \left(\frac{e^h-1}{h}\right)\\ \theta & =\frac{1}{h}\log \left(\frac{e^h-1}{h}\right)\end{array}$$

   • f(x)=logx$f\left(x\right)=\log x$

   The given function f(x)$f\left(x\right)$ is continuous in [x,x+h]$[x,x+h]$ and differentiable in (x,x+h)$(x,x+h)$. There must exist a value θ$\theta$ such that f(x+h)=f(x)+hf′(x+θh)$f(x+h)=f\left(x\right)+h{f}^{\prime }(x+\theta h)$ where 0<θ<1$0<\theta <1$. This is the alternate form of Lagrange’s MVT.

   f(x)=logxf(x+h)=log(x+h)f′(x)=1xf′(x+θh)=1x+θh$$\begin{array}{cc}f\left(x\right)& =\log x\\ f(x+h)& =\log (x+h)\\ f^{\prime }\left(x\right)& =\frac{1}{x}\\ f^{\prime }(x+\theta h)& =\frac{1}{x+\theta h}\end{array}$$

   Substituting these values in alternate form of Lagrange’s MVT, we have,

   log(x+h)=logx+h1x+θhx+θh=hlog(x+h)−logxθ=1h(hlogx+hx−x)θ=1log(1+hx)−xh

7.1.5 Question 5

If in Cauchy’s Mean Value Theorem

1. f(x)=x2 and F(x)=x in [−1,3] show that c∈[−1,3] is the arithmetic mean of −1 and 3.

The two given functions are continuous in [−1,3] and derivable in (−1,3), so the conditions of Cauchy’s MVT are satisfied. Using Cauchy’s MVT, we have,

f(b)−f(a)F(b)−F(a)=f′(c)F′(c)9−13+1=2c12c=2c=1

which is the arithmetic mean of −1 and 3 and lies in (−1,3).

2. f(x)=√x and F(x)=1√x in [a,b], show that c∈[a,b] is the geometric mean of a and b.

The two given functions are continuous in [a,b] and derivable in (a,b), so the conditions of Cauchy’s MVT are satisfied. Using Cauchy’s MVT, we have,

f(b)−f(a)F(b)−F(a)=f′(c)F′(c)√b−√a1√b−1√a=12√c−12c3/2√b−√a−(√b−√a)√ab=−1√c×c3/2−√ab=−cc=√ab

which is the geometric mean of a and b and lies in interval (a,b).

7.1.6 Question 6

1. Examine the validity of Rolle’s theorem for the function f(x)=|x−1| in −1≤x≤3.

The plotting of the function reveals this graph:

Figure 7.3: Graph of f(x)=|x−1|{−1≤x≤3}

From the graph, though the function f(x) is continuous in the interval [−1,3], it does not have derivative at x=1. In other words, right-hand derivative ≠ left-hand derivative at point x=1. Thus the function is not differentiable in the interval [−1,3]. And hence Rolle’s theorem cannot be verified for the given interval.

2. Explain the failure of Lagrange’s Mean Value Theorem for f(x)=2+(x−1)2/3 in [0,2].

The given function is not a polynomial because the degree of equation is a rational number, not a whole number.

The plotting reveals this graph,

Figure 7.4: Graph of f(x)=2+(x−1)2/3{0≤x≤2}

Finding the derivative of the equation,

f′(x)=0+23(x−1)23−1=23(x−1)1/3

At point x=1 in the given open interval (0,2), f′(x)=∞. Thus f′(x) does not exist at (0,2). Hence one of the conditions of Lagrange’s MVT is violated. Hence the failure.

3. Can Cauchy’s Mean Value theorem be applied for the pair f(x)=x2 and F(x)=x3 in the interval [−2,3].

Both the functions are polynomials, are continuous in [−2,3] and derivable in (−2,3). Now, using Cauchy’s MVT, we have,

f(b)−f(a)F(b)−F(a)=f′(c)F′(c)32−(−2)233−(−2)3=2c3c2535=23cc=143

The value of c does not lie in interval (−2,3). Hence, Cauchy’s MVT cannot be applied.