For the functionF(x)=−x4+2x2+8,complete the following parts.(a) Determine whether F is even, odd, or neither.(b) There is a local maximum value of9atx=1.Determine a second local maximum.(c) Suppose the area under the graph of F betweenx=0andx=2that is bounded from below by the x-axis is14.9square units. Using the result from part (a), determine the area under the graph of F betweenx=−2andx=0that is bounded from below by the x-axis.

Answered: For the function F(x)=−x4+2x2+8,…

College Algebra

1st Edition

ISBN:9781938168383

Author:Jay Abramson

Publisher:Jay Abramson

Chapter3: Functions

Section3.3: Rates Of Change And Behavior Of Graphs

Problem 23SE: For the following exercises, consider the graph shown in Figure 16. Estimate the point(s) at which...

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Related questions

Question

For the function

F(x)=−x4+2x2+8,

complete the following parts.

(a) Determine whether F is even, odd, or neither.

(b) There is a local maximum value of

x=1.

Determine a second local maximum.

x=0

and

x=2

that is bounded from below by the x-axis is

14.9

square units. Using the result from part (a), determine the area under the graph of F between

x=−2

and

x=0

that is bounded from below by the x-axis.

Expert Solution

Step 1

$\begin{array}{l} P r o v i d e d f u n c t i o n i s \\ f (x) = - x^{4} + 2 x^{2} + 8 \\ (a) S i n c e w e k n o w t h a t i f f (x) i s e v e n t h e n f (x) = f (- x) . S o l e t u s c h e c k \\ f o r g i v e n f u n c t i o n . \\ f (- x) = - (- x)^{4} + 2 (- x)^{2} + 8 \\ = - x^{4} + 2 x^{2} + 8 \\ = f (x) \\ ⟹ f (- x) = f (x) \\ t h a t i s, g i v e n f u n c t i o n i s E v e n f u n c t i o n . \\ (b) f (x) = - x^{4} + 2 x^{2} + 8 t h e n f' (x) = - 4 x^{3} + 4 x \\ F o r c r i t i c a l p o i n t s p u t f' (x) = 0 \\ - 4 x^{3} + 4 x = 0 \\ - 4 x (x^{2} - 1) = 0 \\ x (x - 1) (x + 1) = 0 \\ x = 0, - 1, 1 \\ N o w e v a l u a t e f'' (x) t o c h e c k m a x / m i n s o \\ f'' (x) = - 12 x^{2} + 4 \\ A t x = - 1, 1, f'' (x) = - 12 + 4 = - 8 \\ f'' (- 1) < 0 a n d f'' (1) < 0 \\ H e n c e f (x) a t t e n d s m a x i m u m a t x = - 1, 1 \\ N o w a t x = 0, f'' (0) = 4 > 0, f (x) a t t e n d s m i n i m u m a t x = 0 . \\ T h e n s e c o n d l o c a l m a x i m u m w i l l b e a t x = - 1, t h e n v a l u e i s \\ f (- 1) = - (- 1)^{4} + 2 (- 1)^{2} + 8 = - 1 + 2 + 8 = 9 \\ s e c o n d l o c a l v a l u e i s a l s o 9 . \end{array}$