Math

QuestionFor the function F(x)=x4+4x2+192F(x)=-x^{4}+4 x^{2}+192, find if FF is even/odd, a second local max, and area from x=4x=-4 to x=0x=0.

Studdy Solution

STEP 1

Assumptions1. The function is (x)=x4+4x+192(x)=-x^{4}+4 x^{}+192 . We need to determine if the function is even, odd, or neither3. There is a local maximum value of196 at x=x=\sqrt{}
4. We need to find a second local maximum value5. The area of the region enclosed by the graph of and the $x$-axis between $x=0$ and $x=4$ is648.5 square units6. We need to find the area of the region enclosed by the graph of and the xx-axis between x=4x=-4 and x=0x=0

STEP 2

To determine if a function is even, odd, or neither, we can use the following properties- A function (x)(x) is even if (x)=F(x)(-x) = F(x) for all xx in the domain of - A function $(x)$ is odd if $(-x) = -(x)$ for all $x$ in the domain of
Let's substitute x-x into the function and simplify.
(x)=(x)4+4(x)2+192(-x) = -(-x)^{4}+4 (-x)^{2}+192

STEP 3

implify the equation.
(x)=x+x2+192(-x) = -x^{}+ x^{2}+192

STEP 4

Since (x)=F(x)(-x) = F(x), we can conclude that the function (x)(x) is even.

STEP 5

The function (x)(x) has a local maximum at x=2x=\sqrt{2}. Since (x)(x) is an even function, it is symmetric about the y-axis. Therefore, there is another local maximum at x=2x=-\sqrt{2}.

STEP 6

To find the value of the second local maximum, we substitute x=2x=-\sqrt{2} into the function.
(2)=(2)4+4(2)2+192(-\sqrt{2}) = -(-\sqrt{2})^{4}+4 (-\sqrt{2})^{2}+192

STEP 7

implify the equation.
(2)=22+42+192=196(-\sqrt{2}) = -2^{2}+4 \cdot2+192 =196So, the second local maximum value is196.

STEP 8

The area of the region enclosed by the graph of and the $x$-axis between $x=0$ and $x=4$ is given as648.5 square units. Since $(x)$ is an even function, the area of the region enclosed by the graph of and the xx-axis between x=4x=-4 and x=0x=0 is also648.5 square units.

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