Question

Determine whether the following functions
i). f(x) = 2x3 - 3x + 1; [-2; 2]
ii). f(x) = e^x, [0; log 4]
iii). f(x) = log 2x, [1; e]
iv). f(x) = sin^(-1) x, [0; 1/2]
meet the conditions of the Mean Value Theorem on the interval. If so, nd the
point(s) guaranteed to exist by the theorem

Answer 1

i) $f(x)=2x^3-3x+1$ is continuous and differentiable for $x\in \R$ as polynomial.

Then $f$ meets the conditions of the Mean Value Theorem on the interval $[-2, 2]$

$f'(x)=6x^2-3=>f(c)=6c^2-3$

$f(-2)=2(-2)^3-3(-2)+1=-9$

$f(2)=2(2)^3-3(2)+1=11$

$$11-(-9)=(6c^2-3)(2-(-2))$$

$$6c^2-3=5$$

$$c^2=\dfrac{4}{3}$$

$$c_1=-\dfrac{2\sqrt{3}}{3}, c_2=\dfrac{2\sqrt{3}}{3}$$

ii) $f(x)=e^x$ is continuous and differentiable for $x\in \R$ .

Then $f$ meets the conditions of the Mean Value Theorem on the interval $[0, \log4].$

Let $\log$ denotes natural logarithm.

$f'(x)=e^x=>f(c)=e^c$

$f(0)=e^0=1$

$f(\log4)=e^{\log4}=4$

$$4-1=e^c(\log4-1)$$

$$e^c=\dfrac{3}{\log4-1}$$

$$c=\log(\dfrac{3}{\log4-1})$$

iii) $f(x)=\log_2(x)$ is continuous and differentiable for $x>0$ .

Then $f$ meets the conditions of the Mean Value Theorem on the interval $[1, e].$

$f'(x)=\dfrac{1}{x\ln2}=>f(c)=\dfrac{1}{c\ln2}$

$f(1)=\log_2(1)=0$

$f(e)=\log_2(e)=\dfrac{1}{\ln2}$

$$\dfrac{1}{\ln2}-0=\dfrac{1}{c\ln2}(e-1)$$

$$c=e-1$$

iv) $f(x)=\sin^{-1}x$ is continuous on $[0, 1/2]$ and differentiable on $(0,1/2).$

Then $f$ meets the conditions of the Mean Value Theorem on the interval $[0, 1/2].$

$f'(x)=\dfrac{1}{\sqrt{1-x^2}}=>f(c)=\dfrac{1}{\sqrt{1-c^2}}$

$f(0)=0$

$f(1/2)=\dfrac{\pi}{6}$

$$\dfrac{\pi}{6}-0=\dfrac{1}{\sqrt{1-c^2}}(\dfrac{1}{2}-0)$$

$$\sqrt{1-c^2}=\dfrac{3}{\pi}$$

$$c^2=1-\dfrac{9}{\pi^2}, c>0$$

$$c=\dfrac{\sqrt{\pi^2-9}}{\pi}$$