Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 4 - Applications of the Derivative - 4.4 The Shape of a Graph - Exercises - Page 195: 23


Critical points: x=3 and x=5. Local maximum value= f(3)= 54. Local minimum value= f(5)= 50.

Work Step by Step

$f'(x)=3x^{2}-24x+45$ $=3(x^{2}-8x+15)=3(x-3)(x-5)$ $f'(x)=0$ at x= 3 and x=5. That is, x=3 and x=5 are the critical points. Now, $f''(x)= 6x-24$ $f''(3)= -6$ $f''(5)= 6$ Therefore, by second derivative test, x=3 is a point of local maxima and local maximum value of f at x=3 is $f(3)= 3^{3}-(12\times3^{2})+(45\times3)=54$ while x=5 is the point of local minima and local minimum value of f at x=5 is $f(5)= 5^{3}-(12\times5^{2})+(45\times5)= 50$
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