Solved on Jan 24, 2024

Find the relative maximum point of $f(x) = -x^3 + 3x^2 - 4$ using $f'(x)$ and $f''(x)$ .

STEP 1

Assumptions
1. The function given is $f(x) = -x^3 + 3x^2 - 4$ .
2. The first derivative of the function is $f'(x) = -3x^2 + 6x$ .
3. The second derivative of the function is $f''(x) = -6x + 6$ .
4. A relative maximum point occurs where the first derivative $f'(x)$ is zero and the second derivative $f''(x)$ is negative.

STEP 2

To find the critical points where $f(x)$ could have a relative maximum, we need to set the first derivative $f'(x)$ equal to zero and solve for $x$ .
$f'(x) = 0$

STEP 3

Substitute the expression for $f'(x)$ and solve for $x$ .
$-3x^2 + 6x = 0$

STEP 4

Factor out the common term $x$ .
$x(-3x + 6) = 0$

STEP 5

Set each factor equal to zero and solve for $x$ .
$x = 0 \quad \text{or} \quad -3x + 6 = 0$

STEP 6

Solve the second equation for $x$ .
$-3x + 6 = 0 \Rightarrow -3x = -6 \Rightarrow x = 2$

STEP 7

Now we have two critical points, $x = 0$ and $x = 2$ . We need to determine which, if any, of these points is a relative maximum. To do this, we will use the second derivative test.

STEP 8

Evaluate the second derivative $f''(x)$ at each critical point.
First, for $x = 0$ :
$f''(0) = -6(0) + 6 = 6$

STEP 9

Since $f''(0) > 0$ , the point $x = 0$ is not a relative maximum (it could be a relative minimum).

STEP 10

Now, evaluate the second derivative $f''(x)$ at the second critical point $x = 2$ :
$f''(2) = -6(2) + 6 = -6$

STEP 11

Since $f''(2) < 0$ , the point $x = 2$ is a relative maximum.

STEP 12

To find the y-coordinate of the relative maximum point, we substitute $x = 2$ into the original function $f(x)$ .
$f(2) = -(2)^3 + 3(2)^2 - 4$

STEP 13

Calculate the value of $f(2)$ .
$f(2) = -8 + 12 - 4 = 0$

STEP 14

The relative maximum point is therefore at $(x, y) = (2, 0)$ .
The relative maximum point is $(2, 0)$ .

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Find the relative maximum point of $f(x) = -x^3 + 3x^2 - 4$ using $f'(x)$ and $f''(x)$ .

STEP 1

STEP 2

To find the critical points where $f(x)$ could have a relative maximum, we need to set the first derivative $f'(x)$ equal to zero and solve for $x$ .
$f'(x) = 0$

STEP 3

Substitute the expression for $f'(x)$ and solve for $x$ .
$-3x^2 + 6x = 0$

STEP 4

Factor out the common term $x$ .
$x(-3x + 6) = 0$

STEP 5

Set each factor equal to zero and solve for $x$ .
$x = 0 \quad \text{or} \quad -3x + 6 = 0$

STEP 6

Solve the second equation for $x$ .
$-3x + 6 = 0 \Rightarrow -3x = -6 \Rightarrow x = 2$

STEP 7

Now we have two critical points, $x = 0$ and $x = 2$ . We need to determine which, if any, of these points is a relative maximum. To do this, we will use the second derivative test.

STEP 8

Evaluate the second derivative $f''(x)$ at each critical point.
First, for $x = 0$ :
$f''(0) = -6(0) + 6 = 6$

STEP 9

Since $f''(0) > 0$ , the point $x = 0$ is not a relative maximum (it could be a relative minimum).

STEP 10

Now, evaluate the second derivative $f''(x)$ at the second critical point $x = 2$ :
$f''(2) = -6(2) + 6 = -6$

STEP 11

Since $f''(2) < 0$ , the point $x = 2$ is a relative maximum.

STEP 12

To find the y-coordinate of the relative maximum point, we substitute $x = 2$ into the original function $f(x)$ .
$f(2) = -(2)^3 + 3(2)^2 - 4$

STEP 13

Calculate the value of $f(2)$ .
$f(2) = -8 + 12 - 4 = 0$

STEP 14

The relative maximum point is therefore at $(x, y) = (2, 0)$ .
The relative maximum point is $(2, 0)$ .

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Find the relative maximum point of f(x)=−x3+3x2−4f(x) = -x^3 + 3x^2 - 4f(x)=−x3+3x2−4 using f′(x)f'(x)f′(x) and f′′(x)f''(x)f′′(x).

STEP 1

STEP 2

To find the critical points where f(x)f(x)f(x) could have a relative maximum, we need to set the first derivative f′(x)f'(x)f′(x) equal to zero and solve for xxx.f′(x)=0f'(x) = 0f′(x)=0

STEP 3

Substitute the expression for f′(x)f'(x)f′(x) and solve for xxx.−3x2+6x=0-3x^2 + 6x = 0−3x2+6x=0

STEP 4

Factor out the common term xxx.x(−3x+6)=0x(-3x + 6) = 0x(−3x+6)=0

STEP 5

Set each factor equal to zero and solve for xxx.x=0or−3x+6=0x = 0 \quad \text{or} \quad -3x + 6 = 0x=0or−3x+6=0

STEP 6

Solve the second equation for xxx.−3x+6=0⇒−3x=−6⇒x=2-3x + 6 = 0 \Rightarrow -3x = -6 \Rightarrow x = 2−3x+6=0⇒−3x=−6⇒x=2

STEP 7

Now we have two critical points, x=0x = 0x=0 and x=2x = 2x=2. We need to determine which, if any, of these points is a relative maximum. To do this, we will use the second derivative test.

STEP 8

Evaluate the second derivative f′′(x)f''(x)f′′(x) at each critical point.First, for x=0x = 0x=0:f′′(0)=−6(0)+6=6f''(0) = -6(0) + 6 = 6f′′(0)=−6(0)+6=6

STEP 9

Since f′′(0)>0f''(0) > 0f′′(0)>0, the point x=0x = 0x=0 is not a relative maximum (it could be a relative minimum).

STEP 10

Now, evaluate the second derivative f′′(x)f''(x)f′′(x) at the second critical point x=2x = 2x=2:f′′(2)=−6(2)+6=−6f''(2) = -6(2) + 6 = -6f′′(2)=−6(2)+6=−6

STEP 11

Since f′′(2)<0f''(2) < 0f′′(2)<0, the point x=2x = 2x=2 is a relative maximum.

STEP 12

To find the y-coordinate of the relative maximum point, we substitute x=2x = 2x=2 into the original function f(x)f(x)f(x).f(2)=−(2)3+3(2)2−4f(2) = -(2)^3 + 3(2)^2 - 4f(2)=−(2)3+3(2)2−4

STEP 13

Calculate the value of f(2)f(2)f(2).f(2)=−8+12−4=0f(2) = -8 + 12 - 4 = 0f(2)=−8+12−4=0

STEP 14

The relative maximum point is therefore at (x,y)=(2,0)(x, y) = (2, 0)(x,y)=(2,0).The relative maximum point is (2,0)(2, 0)(2,0).

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Find the relative maximum point of $f(x) = -x^3 + 3x^2 - 4$ using $f'(x)$ and $f''(x)$ .

To find the critical points where $f(x)$ could have a relative maximum, we need to set the first derivative $f'(x)$ equal to zero and solve for $x$ .
$f'(x) = 0$

Substitute the expression for $f'(x)$ and solve for $x$ .
$-3x^2 + 6x = 0$

Factor out the common term $x$ .
$x(-3x + 6) = 0$

Set each factor equal to zero and solve for $x$ .
$x = 0 \quad \text{or} \quad -3x + 6 = 0$

Solve the second equation for $x$ .
$-3x + 6 = 0 \Rightarrow -3x = -6 \Rightarrow x = 2$

Now we have two critical points, $x = 0$ and $x = 2$ . We need to determine which, if any, of these points is a relative maximum. To do this, we will use the second derivative test.

Evaluate the second derivative $f''(x)$ at each critical point.
First, for $x = 0$ :
$f''(0) = -6(0) + 6 = 6$

Since $f''(0) > 0$ , the point $x = 0$ is not a relative maximum (it could be a relative minimum).

Now, evaluate the second derivative $f''(x)$ at the second critical point $x = 2$ :
$f''(2) = -6(2) + 6 = -6$

Since $f''(2) < 0$ , the point $x = 2$ is a relative maximum.

To find the y-coordinate of the relative maximum point, we substitute $x = 2$ into the original function $f(x)$ .
$f(2) = -(2)^3 + 3(2)^2 - 4$

Calculate the value of $f(2)$ .
$f(2) = -8 + 12 - 4 = 0$

The relative maximum point is therefore at $(x, y) = (2, 0)$ .
The relative maximum point is $(2, 0)$ .