Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.3 The Chain Rule - 4.3 Exercises - Page 225: 49

Answer

$x=1$ , $x=3$.

Work Step by Step

The slope of the tangent at x=a is the derivative of f(x), calculated in x=a. The tangent line has slope $f^{\prime}(a)$, and passes through $(a,f(a)).$ \begin{align*} f(x)&=(x^{3}-6x^{2}+9x+1)^{1/2} \quad \\&\text{.. apply chain rule...}& \\ f^{\prime}(x)& =\displaystyle \frac{1}{2}(x^{3}-6x^{2}+9x+1)^{-1/2}\cdot(x^{3}-6x^{2}+9x+1)^\prime & \\ & =\displaystyle \frac{1}{2}(x^{3}-6x^{2}+9x+1)^{-1/2}\cdot(3x^{2}-12x+9) & \\ & =\displaystyle \frac{3(x^{2}-4x+3)}{2\sqrt{x^{3}-6x^{2}+9x+1}}\quad & \\ \end{align*} The derivative is zero when \begin{align*} 3(x^{2}-4x+3)&=0 \quad & \\ x^{2}-4x+3&=0 \quad & \\ & \text{ ...factoring, } & \\ (x-1)(x-3)&=0 \quad & \\ x&=1, \quad x=3 \end{align*} The tangent line is horizontal at $x=1$ and $x=3$.
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