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Chapter 5: Problem 93

(a) Show that \(f(x)=2 x^{3}+3 x^{2}-36 x\) is not one-to-one on \((-\infty, \infty) .\) (b) Determine the greatest value \(c\) such that \(f\) is one-to-one on \((-c, c) .\)

Short Answer

Expert verified
The function \(f(x)=2 x^{3}+3 x^{2}-36 x\) is not one-to-one on \(- \infty, \infty\) because \(f(0) = f(-3)\). The greatest value \(c\) such that the function is one-to-one on \((-c, c)\) is \(c = 3\).

Step by step solution

01

Show that the function is not one-to-one

This can be proven by finding two different values \(a\) and \(b\) such that \(f(a) = f(b)\). If such values exist, this demonstrates that the function is not one-to-one. By inspecting the function \(f(x) = 2x^3+ 3x^2 - 36x\), it can be observed that \(f(0) = f(-3) = 0\). Thus, \(f\) is not one-to-one on \(-\infty, \infty\).
02

Calculate Derivative of the function

The next step is to take the derivative of the function \(f(x)\). The derivative \(f'(x) = 6x^2 + 6x - 36\). This will be used in the next step to find the maximum and minimum points of the function.
03

Find values of \(x\) where the derivative is zero or does not exist

Set the derivative equal to zero and solve for \(x\), \(6x^2 + 6x - 36 = 0\). After solving the equation, we find that \(x_1 = -3\) and \(x_2 = 2\) are the solutions.
04

Determine the interval for one-to-one

The function will be one-to-one between the turning points, and hence the required interval is \(-3 < x < 2\), so \(c=3\). Since we have to maximize the interval and \(x=-3\) and \(x=2\) are the effective boundaries, \(c\) will be the greatest number among \(3\) and \(2\). Therefore, \(c=3\) is the maximum value that the function can be one-to-one on the interval \((-c,c)\).

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