Solved on Dec 14, 2023

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

STEP 1

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

STEP 2

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

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.
$f'(x) = e^{1-x^2} \cdot \frac{d}{dx}(1-x^2)$

STEP 4

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

STEP 5

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

STEP 6

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

STEP 7

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

STEP 8

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

STEP 9

Solve for $x$ .
$x = 0$

STEP 10

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

STEP 11

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

STEP 12

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

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.
$f''(x) = \frac{d}{dx}(-2x) \cdot e^{1-x^2} + (-2x) \cdot \frac{d}{dx}(e^{1-x^2})$

STEP 14

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

STEP 15

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

STEP 16

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

STEP 17

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

STEP 18

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

STEP 19

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

STEP 20

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

STEP 21

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

STEP 22

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

STEP 23

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

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Find ccc such that f′(c)=0f'(c)=0f′(c)=0 for f(x)=e1−x2f(x)=e^{1-x^2}f(x)=e1−x2 and determine if f(x)f(x)f(x) has a local extremum at x=cx=cx=c.

STEP 1

Assumptions1. The function given is f(x)=e1−x2 f(x) = e^{1-x^2} f(x)=e1−x2.2. We need to find the value of c c c such that the derivative of f(x) f(x) f(x), denoted as f′(x) f'(x) f′(x), is equal to 0.3. We need to determine if f(x) f(x) f(x) has a local extremum at x=c x = c x=c.

STEP 2

To find the value of c c c such that f′(c)=0 f'(c) = 0 f′(c)=0, we first need to find the derivative of f(x) f(x) f(x).f′(x)=ddxe1−x2 f'(x) = \frac{d}{dx} e^{1-x^2} f′(x)=dxd​e1−x2

STEP 3

STEP 4

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

STEP 5

Substitute the derivative of the inner function into the expression for f′(x) f'(x) f′(x).f′(x)=e1−x2⋅(−2x) f'(x) = e^{1-x^2} \cdot (-2x) f′(x)=e1−x2⋅(−2x)

STEP 6

Simplify the expression for the derivative.f′(x)=−2xe1−x2 f'(x) = -2xe^{1-x^2} f′(x)=−2xe1−x2

STEP 7

To find the value of c c c such that f′(c)=0 f'(c) = 0 f′(c)=0, we set the derivative equal to zero and solve for x x x.−2xe1−x2=0 -2xe^{1-x^2} = 0 −2xe1−x2=0

STEP 8

Notice that the exponential function e1−x2 e^{1-x^2} e1−x2 is always positive for all real x x x, so it cannot be zero. Therefore, the only way for the product to be zero is if the other factor, −2x -2x −2x, is zero.−2x=0 -2x = 0 −2x=0

STEP 9

Solve for x x x.x=0 x = 0 x=0

STEP 10

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

STEP 11

To determine if f(x) f(x) f(x) has a local extremum at x=c x = c x=c, we need to examine the second derivative of f(x) f(x) f(x) at x=c x = c x=c.f′′(x)=d2dx2e1−x2 f''(x) = \frac{d^2}{dx^2} e^{1-x^2} f′′(x)=dx2d2​e1−x2

STEP 12

STEP 13

STEP 14

Differentiate −2x -2x −2x and e1−x2 e^{1-x^2} e1−x2 separately.ddx(−2x)=−2 \frac{d}{dx}(-2x) = -2 dxd​(−2x)=−2 ddx(e1−x2)=e1−x2⋅(−2x) \frac{d}{dx}(e^{1-x^2}) = e^{1-x^2} \cdot (-2x) dxd​(e1−x2)=e1−x2⋅(−2x)

STEP 15

Substitute the derivatives back into the expression for f′′(x) f''(x) f′′(x).f′′(x)=(−2)⋅e1−x2+(−2x)⋅(e1−x2⋅(−2x)) f''(x) = (-2) \cdot e^{1-x^2} + (-2x) \cdot (e^{1-x^2} \cdot (-2x)) f′′(x)=(−2)⋅e1−x2+(−2x)⋅(e1−x2⋅(−2x))

STEP 16

Simplify the expression for f′′(x) f''(x) f′′(x).f′′(x)=−2e1−x2+4x2e1−x2 f''(x) = -2e^{1-x^2} + 4x^2e^{1-x^2} f′′(x)=−2e1−x2+4x2e1−x2

STEP 17

Factor out the common term e1−x2 e^{1-x^2} e1−x2.f′′(x)=e1−x2(−2+4x2) f''(x) = e^{1-x^2}(-2 + 4x^2) f′′(x)=e1−x2(−2+4x2)

STEP 18

Evaluate the second derivative at x=c x = c x=c, which we found to be 0.f′′(0)=e1−02(−2+4⋅02) f''(0) = e^{1-0^2}(-2 + 4 \cdot 0^2) f′′(0)=e1−02(−2+4⋅02)

STEP 19

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

STEP 20

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

STEP 21

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

STEP 22

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

STEP 23

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

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

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

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

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

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

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

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

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

Solve for $x$ .
$x = 0$

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

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

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

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

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

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

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

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

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

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

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

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