QuestionFor the function $g(x)=-x^{3}+48 x$ , check if it's even, odd, or neither, and find the local maximum value given a local minimum of -128 at -4.

Studdy Solution

STEP 1

Assumptions1. The function is $g(x)=-x^{3}+48x$ . A function is even if $g(-x)=g(x)$ for all $x$ in the domain of $g$
3. A function is odd if $g(-x)=-g(x)$ for all $x$ in the domain of $g$
4. A local maximum value is the highest point in a certain interval of a function.

STEP 2

First, we need to determine if the function is even, odd, or neither. Let's start by finding $g(-x)$ .
$g(-x) = -(-x)^{} +48(-x)$

STEP 3

implify the expression.
$g(-x) = -x^{3} -48x$

STEP 4

Now, we compare $g(-x)$ with $g(x)$ and $-g(x)$ to determine if the function is even, odd, or neither.
$g(x) = -x^{3} +48x$ $-g(x) = x^{3} -48x$

STEP 5

By comparing, we can see that $g(-x)$ is not equal to $g(x)$ and $g(-x)$ is not equal to $-g(x)$ . Therefore, the function is neither even nor odd.

STEP 6

Now, we need to find the local maximum value of the function. We know that the local minimum value of the function is -128 at x = -4. The local maximum value will occur at the x-value that is the opposite of the x-value at the local minimum. So, the x-value at the local maximum is4.

STEP 7

Substitute x =4 into the function to find the local maximum value.
$g(4) = -(4)^{3} +48(4)$

STEP 8

implify the expression to find the local maximum value.
$g(4) = -64 +192 =128$ So, the local maximum value of the function is128.

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QuestionFor the function $g(x)=-x^{3}+48 x$ , check if it's even, odd, or neither, and find the local maximum value given a local minimum of -128 at -4.

Studdy Solution

STEP 1

Assumptions1. The function is $g(x)=-x^{3}+48x$ . A function is even if $g(-x)=g(x)$ for all $x$ in the domain of $g$
3. A function is odd if $g(-x)=-g(x)$ for all $x$ in the domain of $g$
4. A local maximum value is the highest point in a certain interval of a function.

STEP 2

First, we need to determine if the function is even, odd, or neither. Let's start by finding $g(-x)$ .
$g(-x) = -(-x)^{} +48(-x)$

STEP 3

implify the expression.
$g(-x) = -x^{3} -48x$

STEP 4

Now, we compare $g(-x)$ with $g(x)$ and $-g(x)$ to determine if the function is even, odd, or neither.
$g(x) = -x^{3} +48x$ $-g(x) = x^{3} -48x$

STEP 5

By comparing, we can see that $g(-x)$ is not equal to $g(x)$ and $g(-x)$ is not equal to $-g(x)$ . Therefore, the function is neither even nor odd.

STEP 6

Now, we need to find the local maximum value of the function. We know that the local minimum value of the function is -128 at x = -4. The local maximum value will occur at the x-value that is the opposite of the x-value at the local minimum. So, the x-value at the local maximum is4.

STEP 7

Substitute x =4 into the function to find the local maximum value.
$g(4) = -(4)^{3} +48(4)$

STEP 8

implify the expression to find the local maximum value.
$g(4) = -64 +192 =128$ So, the local maximum value of the function is128.

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QuestionFor the function g(x)=−x3+48xg(x)=-x^{3}+48 xg(x)=−x3+48x, check if it's even, odd, or neither, and find the local maximum value given a local minimum of -128 at -4.

Studdy Solution

STEP 1

STEP 2

First, we need to determine if the function is even, odd, or neither. Let's start by finding g(−x)g(-x)g(−x).g(−x)=−(−x)+48(−x)g(-x) = -(-x)^{} +48(-x)g(−x)=−(−x)+48(−x)

STEP 3

implify the expression.g(−x)=−x3−48xg(-x) = -x^{3} -48xg(−x)=−x3−48x

STEP 4

Now, we compare g(−x)g(-x)g(−x) with g(x)g(x)g(x) and −g(x)-g(x)−g(x) to determine if the function is even, odd, or neither.g(x)=−x3+48xg(x) = -x^{3} +48xg(x)=−x3+48x−g(x)=x3−48x-g(x) = x^{3} -48x−g(x)=x3−48x

STEP 5

By comparing, we can see that g(−x)g(-x)g(−x) is not equal to g(x)g(x)g(x) and g(−x)g(-x)g(−x) is not equal to −g(x)-g(x)−g(x). Therefore, the function is neither even nor odd.

STEP 6

Now, we need to find the local maximum value of the function. We know that the local minimum value of the function is -128 at x = -4. The local maximum value will occur at the x-value that is the opposite of the x-value at the local minimum. So, the x-value at the local maximum is4.

STEP 7

Substitute x =4 into the function to find the local maximum value.g(4)=−(4)3+48(4)g(4) = -(4)^{3} +48(4)g(4)=−(4)3+48(4)

STEP 8

implify the expression to find the local maximum value.g(4)=−64+192=128g(4) = -64 +192 =128g(4)=−64+192=128So, the local maximum value of the function is128.

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QuestionFor the function $g(x)=-x^{3}+48 x$ , check if it's even, odd, or neither, and find the local maximum value given a local minimum of -128 at -4.

First, we need to determine if the function is even, odd, or neither. Let's start by finding $g(-x)$ .
$g(-x) = -(-x)^{} +48(-x)$

implify the expression.
$g(-x) = -x^{3} -48x$

Now, we compare $g(-x)$ with $g(x)$ and $-g(x)$ to determine if the function is even, odd, or neither.
$g(x) = -x^{3} +48x$ $-g(x) = x^{3} -48x$

By comparing, we can see that $g(-x)$ is not equal to $g(x)$ and $g(-x)$ is not equal to $-g(x)$ . Therefore, the function is neither even nor odd.

Substitute x =4 into the function to find the local maximum value.
$g(4) = -(4)^{3} +48(4)$

implify the expression to find the local maximum value.
$g(4) = -64 +192 =128$ So, the local maximum value of the function is128.