Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.3 - How Derivatives Affect the Shape of a Graph - 4.3 Exercises: 10

Answer

(a) $f$ increases on $(-\infty, 1)$ and $(2,\infty)$ and decreases on $(1,2)$. (b) $(1,2)$ is the local maximum and $(2,1)$ is the local minimum of $f$. (c) $f$ is concave upward on $(\frac{3}{2},+\infty)$ and concave downward on $(-\infty, \frac{3}{2})$ Point $(\frac{3}{2},\frac{3}{2})$ is the inflection point of $f$.

Work Step by Step

$$f(x)=2x^3-9x^2+12x-3$$ First, we need to find $f'(x)$ and $f''(x)$. $$f'(x)=6x^2-18x+12$$ $$f'(x)=6(x^2-3x+2)$$ $$f'(x)=6(x-1)(x-2)$$ $$f''(x)=12x-18$$ $$f''(x)=6(2x-3)$$ (a) Take $f'(x)=0$ $$6(x-1)(x-2)=0$$ $$x=1\hspace{.5cm}or\hspace{.5cm}x=2$$ These are the critical points of $f$. To find the trend of increase or decrease of $f$, it's all about the sign of $f'$. A rather useful method here is to build the sign table for $f'$ using the critical points for the intervals as the image below. As we can see, on $(-\infty,1)$ and $(2,+\infty)$, $f'(x)\gt0$, so $f$ would be increasing there. On $(1,2)$, $f'(x)\lt0$, so $f$ would be decreasing on this interval. (b) To find local maximum and minimum, we would use the First Derivative Test. Looking at the table again, at $x=1$, $f'$ changes from positive to negative. So $f$ has a local maximum at $x=1$. At $x=2$, $f'$ changes from negative to positive. So $f$ has a local minimum at $x=2$. $$f(1)=2\times1^3-9\times1^2+12\times1-3=2-9+12-3=2$$ $$f(2)=2\times2^3-9\times2^2+12\times2-3=16-36+24-3=1$$ Therefore, $(1,2)$ is the local maximum and $(2,1)$ the local minimum of $f$. (c) To find out the concavity, we take a close look at the change of signs of $f''(x)$. A table for the signs of $f''(x)$ is also below. Using the Concavity Test and from the table, we see that - On $(-\infty, \frac{3}{2})$, $f''(x)\lt0$, so $f$ is concave downward. - On $(\frac{3}{2},+\infty)$, $f''(x)\gt0$, so $f$ is concave upward. Finally, at $x=\frac{3}{2}$, $f$ changes from concave downward to upward. $$f(\frac{3}{2})=2\times(\frac{3}{2})^3-9\times(\frac{3}{2})^2+12\times\frac{3}{2}-3=\frac{27}{4}-\frac{81}{4}+18-3=\frac{3}{2}$$ So, point $(\frac{3}{2},\frac{3}{2})$ is the inflection point of $f$.
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