Question

consider the function ( f(x)=4x^{3}-4x ) on the interval ( -5,5 ). find the average or mean slope of the function on this interval. by the mean value theorem, we know there exists at least one ( c ) in the open interval ( (-5,5) ) such that ( f(c) ) is equal to this mean slope. for this problem, there are two values of ( c ) that work. the smaller one is and the larger one is question help: video message instructor

Explanation:

Step1: Calculate the function values at endpoints

The formula for the mean - slope (average rate of change) of a function \(y = f(x)\) on the interval \([a,b]\) is \(\frac{f(b)-f(a)}{b - a}\).
Given \(a=-5\), \(b = 5\), and \(f(x)=4x^{3}-4x\).
First, find \(f(-5)\):

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Then, find \(f(5)\):

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Step2: Calculate the mean - slope

The mean - slope \(m=\frac{f(5)-f(-5)}{5-(-5)}\).
Substitute \(f(5) = 480\) and \(f(-5)=-480\) into the formula:

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Step3: Find the derivative of the function

Differentiate \(f(x)=4x^{3}-4x\) using the power rule \((x^{n})^\prime=nx^{n - 1}\).
\(f^\prime(x)=(4x^{3}-4x)^\prime=4\times3x^{2}-4=12x^{2}-4\)

Step4: Solve for \(c\)

Since \(f^\prime(c)=96\) (from the Mean - Value Theorem), we set up the equation \(12c^{2}-4 = 96\).
Add \(4\) to both sides: \(12c^{2}=96 + 4=100\).
Divide both sides by \(12\): \(c^{2}=\frac{100}{12}=\frac{25}{3}\).
Take the square root of both sides: \(c=\pm\sqrt{\frac{25}{3}}=\pm\frac{5}{\sqrt{3}}=\pm\frac{5\sqrt{3}}{3}\approx\pm2.89\)

Answer:

The average or mean slope of the function is \(96\). The smaller value of \(c\) is \(-\frac{5\sqrt{3}}{3}\) and the larger value of \(c\) is \(\frac{5\sqrt{3}}{3}\)