Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.3* The Natural Exponential Function - 6.3* Exercises: 76

Answer

(a) $g'(0)=c+5$ and $g''(0)=c^{2}-2$ (b) $y=(5+3k)x+3$

Work Step by Step

As we are given that $g(x)=e^{cx}+f(x)$ and $h(x)=e^{kx}f(x)$ where, $f(0)=3, f'(0)=5$ and $f''(0)=-2$ (a) $g'(x)=ce^{cx}+f'(x)$ $g'(0)=ce^{0}+f'(0)=c+5$ This implies $g'(0)=c+5$ Now, $g''(x)=c^{2}e^{cx}+f''(x)$ $g''(0)=c^{2}e^{0}+f''(0)=c^{2}-2$ Hence, $g'(0)=c+5$ and $g''(0)=c^{2}-2$ (b) Using product rule, we have $h'(x)=e^{kx}f'(x)+ke^{kx}f(x)$ $h'(0)=e^{0}f'(0)+ke^{0}f(0)=5+3k$ The equation of tangent line to the graph of the function of $h(x)=e^{kx}f(x)$ at x = 0 is given as follows: $y-h(0)=h'(0)(x-0)$ $y-3=(5+3k).x$ This implies Hence, $y=(5+3k)x+3$
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