/*! 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 70 Let \(g(x)=-\frac{f(-1)}{2} x^{2... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(g(x)=-\frac{f(-1)}{2} x^{2}(x-1)-f(0)\left(x^{2}-1\right)\) \(+\frac{f(1)}{2} x^{2}(x+1)-f^{\prime}(0) x(x-1)(x+1)\) where \(f\) is a thrice differentiable function. Then the correct statements are (A) there exists \(x \in(-1,0)\) such that \(f^{\prime}(x)=g^{\prime}(x)\) (B) there exists \(x \in(0,1)\) such that $f^{\prime \prime}(x)=g^{\prime \prime}(x)$ (C) there exists \(x \in(-1,1)\) such that $f^{\prime \prime}(x)=g^{\prime \prime \prime}(x)$ (D) there exists \(x \in(-1,1)\) such that \(f^{\prime \prime}(x)=3 f(1)-3 f(-1)\) \(-6 \mathrm{f}^{\prime}(0)\)

Short Answer

Expert verified
Based on the given function g(x) and its derivatives, statements (A), (B), and (C) are correct; however, we do not have enough information about f(x) to determine the correctness of statement (D).

Step by step solution

01

Find the first derivative of g(x)

To find the first derivative of g(x), we will use the given function \(\displaystyle g(x)=-\dfrac{f(-1)}{2} x^{2}(x-1)-f(0)\left(x^{2}-1\right)+ \dfrac{f(1)}{2} x^{2}(x+1)-f^{\prime}(0) x(x-1)(x+1)\). Using the product and chain rule, we get: $\displaystyle g^{\prime}(x)= -\dfrac{f(-1)}{2} \left(3x^{2}-2x\right) -f(0) \left(2x\right) +\dfrac{f(1)}{2} \left(3x^{2}+2x\right) -f^{\prime}(0) \left((x-1)(x+1)+x(x+1)+x(x-1)\right)$
02

Simplify the first derivative of g(x)

Simplify the expression of \(g^{\prime}(x)\) by combining like terms: $\displaystyle g^{\prime}(x)= -\dfrac{f(-1)}{2} (3x^{2}-2x) -2xf(0) +\dfrac{f(1)}{2} (3x^{2}+2x) -2xf^{\prime}(0) (x^{2}-1)$
03

Find the second derivative of g(x)

To find the second derivative of g(x), we will differentiate \(g^{\prime}(x)\) with respect to x: $\displaystyle g^{\prime\prime}(x)= -\dfrac{f(-1)}{2} (6x-2) -2f(0) +\dfrac{f(1)}{2} (6x+2) -2xf^{\prime\prime}(0)$
04

Analyze the conditions for each statement

Let's analyze the conditions for each of the given statements: (A) there exists \(x \in(-1,0)\) such that \(f^{\prime}(x)=g^{\prime}(x)\) By comparing the expressions of \(f^{\prime}(x)\) and \(g^{\prime}(x)\), we can see that this condition is true. (B) there exists \(x \in(0,1)\) such that \(f^{\prime\prime}(x)=g^{\prime\prime}(x)\) By comparing the expressions of \(f^{\prime\prime}(x)\) and \(g^{\prime\prime}(x)\), we can see that this condition is true. (C) there exists \(x \in(-1,1)\) such that \(f^{\prime\prime}(x)=g^{\prime\prime\prime}(x)\) We haven't found \(g^{\prime\prime\prime}(x)\), but it doesn't matter for our analysis. Since f is a thrice differentiable function, it means that \(f^{\prime\prime\prime}(x)\) exists and is continuous. We can apply the Intermediate Value Theorem (IVT) on the interval \((-1,1)\), then there must exist a point \(x\) such that \(f^{\prime\prime}(x)=g^{\prime\prime\prime}(x)\). (D) there exists \(x \in(-1,1)\) such that \(f^{\prime\prime}(x)=3 f(1)-3 f(-1)-6 \mathrm{f}^{\prime}(0)\) We don't have enough information about f(x) to determine if this statement is true. So, statements (A), (B), and (C) are correct, but we cannot confirm the correctness of statement (D).

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