Math / Calculus

QuestionConsider the function $f(x)$ whose second derivative is $f^{\prime \prime}(x)=6 x+3 \sin (x)$ . If $f(0)=4$ and $f^{\prime}(0)=2$ , what is $f(4)$ ?

Studdy Solution

STEP 1

What is this asking? We're given the second derivative of a function and some initial conditions, and we need to find the function's value at $x = 4$ . Watch out! Don't forget to use both initial conditions, for $f(x)$ and $f'(x)$ , when finding the constants of integration!

STEP 2

1. Find the first derivative.
2. Find the original function.
3. Calculate $f(4)$ .

STEP 3

We know $f''(x) = 6x + 3\sin(x)$ .
To find $f'(x)$ , we need to integrate $f''(x)$ with respect to $x$ .
Let's do it term by term!
The integral of $6x$ is $3x^2$ , and the integral of $3\sin(x)$ is $-3\cos(x)$ .
Don't forget the constant of integration, which we'll call $C_1$ .
$f'(x) = \int f''(x) \, dx = \int (6x + 3\sin(x)) \, dx = 3x^2 - 3\cos(x) + C_1$

STEP 4

We're given that $f'(0) = 2$ .
Let's plug in $x = 0$ into our expression for $f'(x)$ :
$f'(0) = 3(0)^2 - 3\cos(0) + C_1 = -3 + C_1$ Since $f'(0) = 2$ , we have $2 = -3 + C_1$ , which means $C_1 = \mathbf{5}$ .

STEP 5

Now we know the complete expression for the first derivative:
$f'(x) = 3x^2 - 3\cos(x) + 5$

STEP 6

To find $f(x)$ , we integrate $f'(x)$ with respect to $x$ .
The integral of $3x^2$ is $x^3$ , the integral of $-3\cos(x)$ is $-3\sin(x)$ , and the integral of $5$ is $5x$ .
We'll add another constant of integration, $C_2$ .
$f(x) = \int f'(x) \, dx = \int (3x^2 - 3\cos(x) + 5) \, dx = x^3 - 3\sin(x) + 5x + C_2$

STEP 7

We're given that $f(0) = 4$ .
Plugging in $x = 0$ into our expression for $f(x)$ gives us:
$f(0) = (0)^3 - 3\sin(0) + 5(0) + C_2 = C_2$ Since $f(0) = 4$ , we have $C_2 = \mathbf{4}$ .

STEP 8

Now we have the full expression for $f(x)$ :
$f(x) = x^3 - 3\sin(x) + 5x + 4$

STEP 9

Finally, we can find $f(4)$ by substituting $x = 4$ into our expression for $f(x)$ :
$f(4) = (4)^3 - 3\sin(4) + 5(4) + 4$

STEP 10

$f(4) = 64 - 3\sin(4) + 20 + 4 = 88 - 3\sin(4)$

STEP 11

$f(4) = 88 - 3\sin(4)$ .

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QuestionConsider the function $f(x)$ whose second derivative is $f^{\prime \prime}(x)=6 x+3 \sin (x)$ . If $f(0)=4$ and $f^{\prime}(0)=2$ , what is $f(4)$ ?

Studdy Solution

STEP 1

What is this asking? We're given the second derivative of a function and some initial conditions, and we need to find the function's value at $x = 4$ . Watch out! Don't forget to use both initial conditions, for $f(x)$ and $f'(x)$ , when finding the constants of integration!

STEP 2

1. Find the first derivative.
2. Find the original function.
3. Calculate $f(4)$ .

STEP 3

STEP 4

We're given that $f'(0) = 2$ .
Let's plug in $x = 0$ into our expression for $f'(x)$ :
$f'(0) = 3(0)^2 - 3\cos(0) + C_1 = -3 + C_1$ Since $f'(0) = 2$ , we have $2 = -3 + C_1$ , which means $C_1 = \mathbf{5}$ .

STEP 5

Now we know the complete expression for the first derivative:
$f'(x) = 3x^2 - 3\cos(x) + 5$

STEP 6

STEP 7

We're given that $f(0) = 4$ .
Plugging in $x = 0$ into our expression for $f(x)$ gives us:
$f(0) = (0)^3 - 3\sin(0) + 5(0) + C_2 = C_2$ Since $f(0) = 4$ , we have $C_2 = \mathbf{4}$ .

STEP 8

Now we have the full expression for $f(x)$ :
$f(x) = x^3 - 3\sin(x) + 5x + 4$

STEP 9

Finally, we can find $f(4)$ by substituting $x = 4$ into our expression for $f(x)$ :
$f(4) = (4)^3 - 3\sin(4) + 5(4) + 4$

STEP 10

$f(4) = 64 - 3\sin(4) + 20 + 4 = 88 - 3\sin(4)$

STEP 11

$f(4) = 88 - 3\sin(4)$ .

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QuestionConsider the function f(x)f(x)f(x) whose second derivative is f′′(x)=6x+3sin⁡(x)f^{\prime \prime}(x)=6 x+3 \sin (x)f′′(x)=6x+3sin(x). If f(0)=4f(0)=4f(0)=4 and f′(0)=2f^{\prime}(0)=2f′(0)=2, what is f(4)f(4)f(4) ?

Studdy Solution

STEP 1

STEP 2

1. Find the first derivative.2. Find the original function.3. Calculate f(4)f(4)f(4).

STEP 3

STEP 4

STEP 5

Now we know the complete expression for the first derivative:f′(x)=3x2−3cos⁡(x)+5f'(x) = 3x^2 - 3\cos(x) + 5f′(x)=3x2−3cos(x)+5

STEP 6

STEP 7

We're given that f(0)=4f(0) = 4f(0)=4.Plugging in x=0x = 0x=0 into our expression for f(x)f(x)f(x) gives us:f(0)=(0)3−3sin⁡(0)+5(0)+C2=C2f(0) = (0)^3 - 3\sin(0) + 5(0) + C_2 = C_2f(0)=(0)3−3sin(0)+5(0)+C2​=C2​Since f(0)=4f(0) = 4f(0)=4, we have C2=4C_2 = \mathbf{4}C2​=4.

STEP 8

Now we have the full expression for f(x)f(x)f(x):f(x)=x3−3sin⁡(x)+5x+4f(x) = x^3 - 3\sin(x) + 5x + 4f(x)=x3−3sin(x)+5x+4

STEP 9

Finally, we can find f(4)f(4)f(4) by substituting x=4x = 4x=4 into our expression for f(x)f(x)f(x):f(4)=(4)3−3sin⁡(4)+5(4)+4f(4) = (4)^3 - 3\sin(4) + 5(4) + 4f(4)=(4)3−3sin(4)+5(4)+4

STEP 10

f(4)=64−3sin⁡(4)+20+4=88−3sin⁡(4)f(4) = 64 - 3\sin(4) + 20 + 4 = 88 - 3\sin(4)f(4)=64−3sin(4)+20+4=88−3sin(4)

STEP 11

f(4)=88−3sin⁡(4)f(4) = 88 - 3\sin(4)f(4)=88−3sin(4).

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QuestionConsider the function $f(x)$ whose second derivative is $f^{\prime \prime}(x)=6 x+3 \sin (x)$ . If $f(0)=4$ and $f^{\prime}(0)=2$ , what is $f(4)$ ?

1. Find the first derivative.
2. Find the original function.
3. Calculate $f(4)$ .

Now we know the complete expression for the first derivative:
$f'(x) = 3x^2 - 3\cos(x) + 5$

We're given that $f(0) = 4$ .
Plugging in $x = 0$ into our expression for $f(x)$ gives us:
$f(0) = (0)^3 - 3\sin(0) + 5(0) + C_2 = C_2$ Since $f(0) = 4$ , we have $C_2 = \mathbf{4}$ .

Now we have the full expression for $f(x)$ :
$f(x) = x^3 - 3\sin(x) + 5x + 4$

Finally, we can find $f(4)$ by substituting $x = 4$ into our expression for $f(x)$ :
$f(4) = (4)^3 - 3\sin(4) + 5(4) + 4$

$f(4) = 64 - 3\sin(4) + 20 + 4 = 88 - 3\sin(4)$

$f(4) = 88 - 3\sin(4)$ .