Math / Calculus

Question6. Find the exact absolute maximum and minimum of $h(x)=x e^{-x^{2}}$ on the interval $[-1,1]$ .
7. Let $f(x)=\ln \left(2 x^{3}-3 x^{2}\right)$ . Find all values of $x$ for which $f^{\prime}(x)$ is 0 or undefined. Determine

Studdy Solution

STEP 1

What is this asking? We need to find the highest and lowest points of a curvy function within a specific range, and then, for a different function, we want to find where its slope is zero or doesn't exist. Watch out! Remember, the edges of the interval can be maximums or minimums too!
Also, don't forget to check where the derivative might not exist, like when the denominator is zero.

STEP 2

1. Find the derivative of $h(x)$ .
2. Find critical points of $h(x)$ .
3. Evaluate $h(x)$ at critical points and endpoints.
4. Find the derivative of $f(x)$ .
5. Find where $f'(x)$ is zero or undefined.

STEP 3

Define the function: $h(x) = x e^{-x^2}$ .
We're going to use the product rule!
Remember, the product rule says $(uv)' = u'v + uv'$ .

STEP 4

Let $u = x$ and $v = e^{-x^2}$ .
Then $u' = 1$ .
To find $v'$ , we need the chain rule!

STEP 5

The chain rule says $(f(g(x)))' = f'(g(x)) \cdot g'(x)$ .
Here, $f(x) = e^x$ and $g(x) = -x^2$ , so $f'(x) = e^x$ and $g'(x) = -2x$ .
Therefore, $v' = e^{-x^2} \cdot (-2x) = -2x e^{-x^2}$ .

STEP 6

Putting it all together with the product rule: $h'(x) = (1) \cdot e^{-x^2} + x \cdot (-2x e^{-x^2}) = e^{-x^2} - 2x^2 e^{-x^2} = e^{-x^2}(1 - 2x^2)$ .
So, $h'(x) = e^{-x^2}(1 - 2x^2)$ is our derivative!

STEP 7

Critical points happen where $h'(x) = 0$ or $h'(x)$ is undefined.
Since $e^{-x^2}$ is never zero and always defined, we only need to look at $1 - 2x^2 = 0$ .

STEP 8

Solving for $x$ , we get $2x^2 = 1$ , so $x^2 = \frac{1}{2}$ , and $x = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$ .
These are our critical points within the interval $[-1, 1]$ .

STEP 9

We need to check $h(x)$ at $x = -1$ , $x = -\frac{\sqrt{2}}{2}$ , $x = \frac{\sqrt{2}}{2}$ , and $x = 1$ .

STEP 10

$h(-1) = (-1)e^{-(-1)^2} = -e^{-1} = -\frac{1}{e}$ . $h\left(-\frac{\sqrt{2}}{2}\right) = \left(-\frac{\sqrt{2}}{2}\right)e^{-\left(-\frac{\sqrt{2}}{2}\right)^2} = -\frac{\sqrt{2}}{2}e^{-\frac{1}{2}} = -\frac{\sqrt{2}}{2\sqrt{e}}$ . $h\left(\frac{\sqrt{2}}{2}\right) = \left(\frac{\sqrt{2}}{2}\right)e^{-\left(\frac{\sqrt{2}}{2}\right)^2} = \frac{\sqrt{2}}{2}e^{-\frac{1}{2}} = \frac{\sqrt{2}}{2\sqrt{e}}$ . $h(1) = (1)e^{-(1)^2} = e^{-1} = \frac{1}{e}$ .

STEP 11

Define the function: $f(x) = \ln(2x^3 - 3x^2)$ .
We'll use the chain rule again!

STEP 12

Let $g(x) = 2x^3 - 3x^2$ .
Then $g'(x) = 6x^2 - 6x$ .
Since the derivative of $\ln(x)$ is $\frac{1}{x}$ , we have $f'(x) = \frac{1}{2x^3 - 3x^2} \cdot (6x^2 - 6x) = \frac{6x^2 - 6x}{2x^3 - 3x^2} = \frac{6x(x-1)}{x^2(2x - 3)}$ .

STEP 13

$f'(x) = 0$ when the numerator is zero: $6x(x-1) = 0$ , so $x = 0$ or $x = 1$ .

STEP 14

$f'(x)$ is undefined when the denominator is zero: $x^2(2x - 3) = 0$ , so $x = 0$ or $2x - 3 = 0$ , meaning $x = \frac{3}{2}$ .

STEP 15

The absolute maximum of $h(x)$ is $\frac{\sqrt{2}}{2\sqrt{e}}$ at $x = \frac{\sqrt{2}}{2}$ , and the absolute minimum is $-\frac{\sqrt{2}}{2\sqrt{e}}$ at $x = -\frac{\sqrt{2}}{2}$ .
For $f'(x)$ , the derivative is zero at $x = 1$ and undefined at $x = 0$ and $x = \frac{3}{2}$ .

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Question6. Find the exact absolute maximum and minimum of $h(x)=x e^{-x^{2}}$ on the interval $[-1,1]$ .
7. Let $f(x)=\ln \left(2 x^{3}-3 x^{2}\right)$ . Find all values of $x$ for which $f^{\prime}(x)$ is 0 or undefined. Determine

Studdy Solution

STEP 1

STEP 2

1. Find the derivative of $h(x)$ .
2. Find critical points of $h(x)$ .
3. Evaluate $h(x)$ at critical points and endpoints.
4. Find the derivative of $f(x)$ .
5. Find where $f'(x)$ is zero or undefined.

STEP 3

Define the function: $h(x) = x e^{-x^2}$ .
We're going to use the product rule!
Remember, the product rule says $(uv)' = u'v + uv'$ .

STEP 4

Let $u = x$ and $v = e^{-x^2}$ .
Then $u' = 1$ .
To find $v'$ , we need the chain rule!

STEP 5

The chain rule says $(f(g(x)))' = f'(g(x)) \cdot g'(x)$ .
Here, $f(x) = e^x$ and $g(x) = -x^2$ , so $f'(x) = e^x$ and $g'(x) = -2x$ .
Therefore, $v' = e^{-x^2} \cdot (-2x) = -2x e^{-x^2}$ .

STEP 6

Putting it all together with the product rule: $h'(x) = (1) \cdot e^{-x^2} + x \cdot (-2x e^{-x^2}) = e^{-x^2} - 2x^2 e^{-x^2} = e^{-x^2}(1 - 2x^2)$ .
So, $h'(x) = e^{-x^2}(1 - 2x^2)$ is our derivative!

STEP 7

Critical points happen where $h'(x) = 0$ or $h'(x)$ is undefined.
Since $e^{-x^2}$ is never zero and always defined, we only need to look at $1 - 2x^2 = 0$ .

STEP 8

Solving for $x$ , we get $2x^2 = 1$ , so $x^2 = \frac{1}{2}$ , and $x = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$ .
These are our critical points within the interval $[-1, 1]$ .

STEP 9

We need to check $h(x)$ at $x = -1$ , $x = -\frac{\sqrt{2}}{2}$ , $x = \frac{\sqrt{2}}{2}$ , and $x = 1$ .

STEP 10

STEP 11

Define the function: $f(x) = \ln(2x^3 - 3x^2)$ .
We'll use the chain rule again!

STEP 12

Let $g(x) = 2x^3 - 3x^2$ .
Then $g'(x) = 6x^2 - 6x$ .
Since the derivative of $\ln(x)$ is $\frac{1}{x}$ , we have $f'(x) = \frac{1}{2x^3 - 3x^2} \cdot (6x^2 - 6x) = \frac{6x^2 - 6x}{2x^3 - 3x^2} = \frac{6x(x-1)}{x^2(2x - 3)}$ .

STEP 13

$f'(x) = 0$ when the numerator is zero: $6x(x-1) = 0$ , so $x = 0$ or $x = 1$ .

STEP 14

$f'(x)$ is undefined when the denominator is zero: $x^2(2x - 3) = 0$ , so $x = 0$ or $2x - 3 = 0$ , meaning $x = \frac{3}{2}$ .

STEP 15

The absolute maximum of $h(x)$ is $\frac{\sqrt{2}}{2\sqrt{e}}$ at $x = \frac{\sqrt{2}}{2}$ , and the absolute minimum is $-\frac{\sqrt{2}}{2\sqrt{e}}$ at $x = -\frac{\sqrt{2}}{2}$ .
For $f'(x)$ , the derivative is zero at $x = 1$ and undefined at $x = 0$ and $x = \frac{3}{2}$ .

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Studdy Solution

STEP 1

STEP 2

1. Find the derivative of h(x)h(x)h(x).2. Find critical points of h(x)h(x)h(x).3. Evaluate h(x)h(x)h(x) at critical points and endpoints.4. Find the derivative of f(x)f(x)f(x).5. Find where f′(x)f'(x)f′(x) is zero or undefined.

STEP 3

**Define the function:** h(x)=xe−x2h(x) = x e^{-x^2}h(x)=xe−x2.We're going to use the **product rule**!Remember, the product rule says (uv)′=u′v+uv′(uv)' = u'v + uv'(uv)′=u′v+uv′.

STEP 4

Let u=xu = xu=x and v=e−x2v = e^{-x^2}v=e−x2.Then u′=1u' = 1u′=1.To find v′v'v′, we need the **chain rule**!

STEP 5

STEP 6

STEP 7

**Critical points** happen where h′(x)=0h'(x) = 0h′(x)=0 or h′(x)h'(x)h′(x) is undefined.Since e−x2e^{-x^2}e−x2 is never zero and always defined, we only need to look at 1−2x2=01 - 2x^2 = 01−2x2=0.

STEP 8

Solving for xxx, we get 2x2=12x^2 = 12x2=1, so x2=12x^2 = \frac{1}{2}x2=21​, and x=±12=±12=±22x = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}x=±21​​=±2​1​=±22​​.These are our **critical points** within the interval [−1,1][-1, 1][−1,1].

STEP 9

We need to check h(x)h(x)h(x) at x=−1x = -1x=−1, x=−22x = -\frac{\sqrt{2}}{2}x=−22​​, x=22x = \frac{\sqrt{2}}{2}x=22​​, and x=1x = 1x=1.

STEP 10

STEP 11

**Define the function:** f(x)=ln⁡(2x3−3x2)f(x) = \ln(2x^3 - 3x^2)f(x)=ln(2x3−3x2).We'll use the **chain rule** again!

STEP 12

STEP 13

f′(x)=0f'(x) = 0f′(x)=0 when the numerator is zero: 6x(x−1)=06x(x-1) = 06x(x−1)=0, so x=0x = 0x=0 or x=1x = 1x=1.

STEP 14

f′(x)f'(x)f′(x) is undefined when the denominator is zero: x2(2x−3)=0x^2(2x - 3) = 0x2(2x−3)=0, so x=0x = 0x=0 or 2x−3=02x - 3 = 02x−3=0, meaning x=32x = \frac{3}{2}x=23​.

STEP 15

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1. Find the derivative of $h(x)$ .
2. Find critical points of $h(x)$ .
3. Evaluate $h(x)$ at critical points and endpoints.
4. Find the derivative of $f(x)$ .
5. Find where $f'(x)$ is zero or undefined.

Define the function: $h(x) = x e^{-x^2}$ .
We're going to use the product rule!
Remember, the product rule says $(uv)' = u'v + uv'$ .

Let $u = x$ and $v = e^{-x^2}$ .
Then $u' = 1$ .
To find $v'$ , we need the chain rule!

Critical points happen where $h'(x) = 0$ or $h'(x)$ is undefined.
Since $e^{-x^2}$ is never zero and always defined, we only need to look at $1 - 2x^2 = 0$ .

Solving for $x$ , we get $2x^2 = 1$ , so $x^2 = \frac{1}{2}$ , and $x = \pm \sqrt{\frac{1}{2}} = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$ .
These are our critical points within the interval $[-1, 1]$ .

We need to check $h(x)$ at $x = -1$ , $x = -\frac{\sqrt{2}}{2}$ , $x = \frac{\sqrt{2}}{2}$ , and $x = 1$ .

Define the function: $f(x) = \ln(2x^3 - 3x^2)$ .
We'll use the chain rule again!

$f'(x) = 0$ when the numerator is zero: $6x(x-1) = 0$ , so $x = 0$ or $x = 1$ .

$f'(x)$ is undefined when the denominator is zero: $x^2(2x - 3) = 0$ , so $x = 0$ or $2x - 3 = 0$ , meaning $x = \frac{3}{2}$ .