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Chapter 2: Problem 66

Use a graphing utility to find the \(x\) -values at which \(f\) is differentiable. \(f(x)=\left\\{\begin{array}{ll}x^{3}-3 x^{2}+3 x, & x \leq 1 \\ x^{2}-2 x, & x>1\end{array}\right.\)

Short Answer

Expert verified
The function \(f(x)\) is differentiable at all \(x\) -values if the function is continuous at \(x=1\) and if the left and right derivatives are equal at \(x=1\).

Step by step solution

01

Define the Functions

The function \(f(x)\) is a piecewise function, which can be broken down into two functions:\[f_{1}(x)=x^{3}-3x^{2}+3x, \qquad x \leq 1\]\[f_{2}(x)=x^{2}-2x, \qquad x > 1\]
02

Calculate the Derivatives

Next, compute the derivative of each function, denoted \(f_{1}'(x)\) and \(f_{2}'(x)\). The derivatives would be:\[f_{1}'(x)= 3x^{2}-6x+3 \]\[f_{2}'(x)= 2x-2 \]
03

Check Continuity at x=1

First, check if the function \(f(x)\) is continuous at \(x=1\) by checking if \[f_{1}(1)=f_{2}(1)\]If the two values are equal, the function is continuous at \(x=1\).
04

Check Differentiability at x=1

If \(f(x)\) is continuous at \(x=1\), then check if it is also differentiable at \(x=1\). This means we need to verify if \[f_{1}'(1)=f_{2}'(1)\]If these two values are equal, the function is differentiable at \(x=1\).
05

Plot the Graph

Finally, to find the \(x\) -values at which \(f\) is differentiable, one would typically plot \(f(x)\), \(f_{1}'(x)\), and \(f_{2}'(x)\) on a graph and observe where \(f(x)\) does not have sharp corners or cusps and where it is smooth.

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