/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 47 Given \(f(x)=4-\left(\frac{1}{2}... [FREE SOLUTION] | 91影视

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Chapter 7: Problem 47

Given \(f(x)=4-\left(\frac{1}{2}-x\right)^{2 / 3} ; g(x)=\\{x\\}\) $h(x)=\left\\{\begin{array}{ll}\frac{\tan [x]}{x}, & x \neq 0 \\ 1 \quad, \quad x=0\end{array}\right.\( and \)p(x)=5^{\ln x+1}\( defined in \)[0,1]$, the functions on which LMVT is applicable is $\begin{array}{ll}\text { (A) } \mathrm{f} & \text { (B) } \mathrm{g}\end{array}$ (C) \(\mathrm{p}\) (D) \(h\)

Short Answer

Expert verified
Answer: (A) f, (B) g, and (D) h

Step by step solution

01

Analyze function f(x)

f(x) is given as \(f(x) = 4 - \left(\frac{1}{2} - x\right)^{\frac{2}{3}}\). This function is continuous since it is a composition of the continuous functions: subtraction, taking the power of \(\frac{2}{3}\), and subtraction of 4. However, let's find its derivative: The derivative of f(x) is given by chain rule: \(f'(x) = -\left(\frac{2}{3}\right) \left(\frac{1}{2}-x\right)^{-\frac{1}{3}} \cdot (-1)\) \(f'(x) = \frac{2}{3} \left(\frac{1}{2}-x\right)^{-\frac{1}{3}}\) Now, let's consider the differentiability of f'(x) on (a,b) = (0,1): The function is differentiable, since its derivative exists for all x in the interval (0,1).
02

Analyze function g(x)

g(x) is given as \(g(x) =\{x\}\). It's equal to the identity function. Since g(x) is a polynomial function, it is continuous and differentiable on (0,1). Thus, it satisfies the conditions for LMVT.
03

Analyze function h(x)

h(x) is given as: $h(x)=\left\{\begin{array}{ll}\frac{\tan [x]}{x}, & x\neq 0 \\\ 1 \quad, \quad x=0\end{array}\right.$. Let's analyze the differentiability at \(x=0\). Using L'H么pital's rule: \(\lim_{x \to 0} \frac{\tan x}{x} = \left[\frac{\tan(x)}{x}\right]' = \frac{\sec^2(x) \cdot x - \tan(x)}{x^2}\). Now, taking the limit: \(\lim_{x \to 0} \frac{\sec^2(x) \cdot x - \tan(x)}{x^2} = \lim_{x \to 0} \frac{1 - 0}{0} = 1\). h(x) is continuous at x=0 and differentiable for all x in (0,1).
04

Analyze function p(x)

p(x) is given as \(p(x) = 5^{\ln x + 1}\) defined on [0,1]. Notice that p(x) is discontinuous at x=0, since \(\ln(0)\) is undefined. Therefore, p(x) does not satisfy the conditions for LMVT on the interval [0,1].
05

Answer

Based on our analysis, the functions that satisfy the conditions for LMVT are: (A) f (B) g (D) h

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