Solved on Dec 14, 2023

Find cc such that f(c)=0f'(c)=0 for f(x)=e1x2f(x)=e^{1-x^2} and determine if f(x)f(x) has a local extremum at x=cx=c.

STEP 1

Assumptions
1. The function given is f(x)=e1x2 f(x) = e^{1-x^2} .
2. We need to find the value of c c such that the derivative of f(x) f(x) , denoted as f(x) f'(x) , is equal to 0.
3. We need to determine if f(x) f(x) has a local extremum at x=c x = c .

STEP 2

To find the value of c c such that f(c)=0 f'(c) = 0 , we first need to find the derivative of f(x) f(x) .
f(x)=ddxe1x2 f'(x) = \frac{d}{dx} e^{1-x^2}

STEP 3

We will use the chain rule for differentiation, which states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function times the derivative of the inner function.
f(x)=e1x2ddx(1x2) f'(x) = e^{1-x^2} \cdot \frac{d}{dx}(1-x^2)

STEP 4

Now we differentiate the inner function 1x2 1-x^2 .
ddx(1x2)=2x \frac{d}{dx}(1-x^2) = -2x

STEP 5

Substitute the derivative of the inner function into the expression for f(x) f'(x) .
f(x)=e1x2(2x) f'(x) = e^{1-x^2} \cdot (-2x)

STEP 6

Simplify the expression for the derivative.
f(x)=2xe1x2 f'(x) = -2xe^{1-x^2}

STEP 7

To find the value of c c such that f(c)=0 f'(c) = 0 , we set the derivative equal to zero and solve for x x .
2xe1x2=0 -2xe^{1-x^2} = 0

STEP 8

Notice that the exponential function e1x2 e^{1-x^2} is always positive for all real x x , so it cannot be zero. Therefore, the only way for the product to be zero is if the other factor, 2x -2x , is zero.
2x=0 -2x = 0

STEP 9

Solve for x x .
x=0 x = 0

STEP 10

We have found that c=0 c = 0 is the value at which f(c)=0 f'(c) = 0 .

STEP 11

To determine if f(x) f(x) has a local extremum at x=c x = c , we need to examine the second derivative of f(x) f(x) at x=c x = c .
f(x)=d2dx2e1x2 f''(x) = \frac{d^2}{dx^2} e^{1-x^2}

STEP 12

We will again use the chain rule to find the second derivative. We already know that f(x)=2xe1x2 f'(x) = -2xe^{1-x^2} , so we differentiate this expression with respect to x x .
f(x)=ddx(2xe1x2) f''(x) = \frac{d}{dx} (-2xe^{1-x^2})

STEP 13

Apply the product rule for differentiation, which states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.
f(x)=ddx(2x)e1x2+(2x)ddx(e1x2) f''(x) = \frac{d}{dx}(-2x) \cdot e^{1-x^2} + (-2x) \cdot \frac{d}{dx}(e^{1-x^2})

STEP 14

Differentiate 2x -2x and e1x2 e^{1-x^2} separately.
ddx(2x)=2 \frac{d}{dx}(-2x) = -2 ddx(e1x2)=e1x2(2x) \frac{d}{dx}(e^{1-x^2}) = e^{1-x^2} \cdot (-2x)

STEP 15

Substitute the derivatives back into the expression for f(x) f''(x) .
f(x)=(2)e1x2+(2x)(e1x2(2x)) f''(x) = (-2) \cdot e^{1-x^2} + (-2x) \cdot (e^{1-x^2} \cdot (-2x))

STEP 16

Simplify the expression for f(x) f''(x) .
f(x)=2e1x2+4x2e1x2 f''(x) = -2e^{1-x^2} + 4x^2e^{1-x^2}

STEP 17

Factor out the common term e1x2 e^{1-x^2} .
f(x)=e1x2(2+4x2) f''(x) = e^{1-x^2}(-2 + 4x^2)

STEP 18

Evaluate the second derivative at x=c x = c , which we found to be 0.
f(0)=e102(2+402) f''(0) = e^{1-0^2}(-2 + 4 \cdot 0^2)

STEP 19

Simplify the expression for f(0) f''(0) .
f(0)=e1(2+0) f''(0) = e^1(-2 + 0)

STEP 20

Calculate f(0) f''(0) .
f(0)=e(2) f''(0) = e \cdot (-2)

STEP 21

Since e e is a positive constant and we have a negative sign, f(0) f''(0) is negative.

STEP 22

A negative second derivative at x=c x = c indicates that f(x) f(x) has a local maximum at x=c x = c .

STEP 23

Therefore, f(x) f(x) has a local extremum at x=c x = c , and specifically, it is a local maximum since f(c)<0 f''(c) < 0 .
The value of c c is 0, and f(x) f(x) has a local maximum at x=c x = c .

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