Math  /  Calculus

Question(1 point) Let f(x)=e7x2f(x)=e^{-7 x^{2}}. Then f(x)f(x) has a relative minimum at x=x= a relative maximum at x=x=\square and inflection points at x=x=\square and at x=x=\square
Write DNE if any of the above do not exist. Write the inflection points (if ar

Studdy Solution

STEP 1

What is this asking? Find where the function f(x)=e7x2f(x) = e^{-7x^2} is lowest and highest nearby, and where its curvature changes. Watch out! Don't mix up minimums and maximums, and remember inflection points are about curvature, not slope!

STEP 2

1. Find the first derivative.
2. Find the critical points.
3. Classify the critical points.
4. Find the second derivative.
5. Find the inflection points.

STEP 3

To find the **critical points**, which tell us about **minimums** and **maximums**, we need the **first derivative**, f(x)f'(x).
Let's use the **chain rule**!

STEP 4

The derivative of eue^u is eududxe^u \cdot \frac{du}{dx}.
Here, u=7x2u = -7x^2, so dudx=14x\frac{du}{dx} = -14x.

STEP 5

Putting it together, we get f(x)=e7x2(14x)=14xe7x2f'(x) = e^{-7x^2} \cdot (-14x) = -14x \cdot e^{-7x^2}.
Awesome!

STEP 6

**Critical points** happen where f(x)=0f'(x) = 0 or f(x)f'(x) is undefined.
Our f(x)f'(x) is defined everywhere!
So, let's set f(x)f'(x) to zero: 14xe7x2=0-14x \cdot e^{-7x^2} = 0.

STEP 7

Since e7x2e^{-7x^2} is *never* zero, we just need 14x=0-14x = 0, which means x=0x = 0.
So, x=0x = 0 is our **only critical point**!

STEP 8

Let's check the sign of f(x)f'(x) around x=0x = 0.
If xx is slightly negative, like 0.1-0.1, then f(x)f'(x) is positive.
If xx is slightly positive, like 0.10.1, then f(x)f'(x) is negative.

STEP 9

Since f(x)f'(x) goes from positive to negative at x=0x = 0, we have a **relative maximum** at x=0x = 0.

STEP 10

**Inflection points** happen where the **second derivative**, f(x)f''(x), changes sign.
Let's find f(x)f''(x) using the **product rule** on f(x)=14xe7x2f'(x) = -14x \cdot e^{-7x^2}.

STEP 11

The **product rule** says (uv)=uv+uv(uv)' = u'v + uv'.
Here, u=14xu = -14x and v=e7x2v = e^{-7x^2}.
We have u=14u' = -14 and v=14xe7x2v' = -14xe^{-7x^2} (from earlier).

STEP 12

So, f(x)=(14)(e7x2)+(14x)(14xe7x2)=(14+196x2)e7x2=14(14x21)e7x2f''(x) = (-14)(e^{-7x^2}) + (-14x)(-14xe^{-7x^2}) = (-14 + 196x^2)e^{-7x^2} = 14(14x^2 - 1)e^{-7x^2}.

STEP 13

We set f(x)=0f''(x) = 0: 14(14x21)e7x2=014(14x^2 - 1)e^{-7x^2} = 0.
Again, e7x2e^{-7x^2} is never zero, so we need 14x21=014x^2 - 1 = 0.

STEP 14

This gives us x2=114x^2 = \frac{1}{14}, so x=±114=±114=±1414x = \pm \sqrt{\frac{1}{14}} = \pm \frac{1}{\sqrt{14}} = \pm \frac{\sqrt{14}}{14}.

STEP 15

These are our **inflection points** because f(x)f''(x) changes sign around them.
If we plug in x=0x=0, we get f(0)=14<0f''(0) = -14 < 0, so the graph is concave down.
For values of xx further away from zero, f(x)f''(x) will be positive, so the graph is concave up.

STEP 16

Relative minimum: DNE Relative maximum: x=0x = 0 Inflection points: x=1414x = \frac{\sqrt{14}}{14} and x=1414x = -\frac{\sqrt{14}}{14}

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