A function has f''(x) = 10 and has f'(4) = 0 and f(2) = 4. Find f(x).

Steve yesterday at 5:55pm

If f" = 10 for all x, then
f' = 10x+c
10*4+c = 0 so c = -40 and f' = 10x-40
f = 5x^2-40x+c
5*2^2-40*2+c = 4 so c=64 and f(x) = 5x^2-40x+64

Oooo -- Jake, Marcus, Sara, Sarah

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To solve this problem, we need to find the antiderivative of f''(x) = 10, which will give us f'(x). Then, we can find the antiderivative of f'(x) to find f(x).

Step 1: Find f'(x)
Since f''(x) = 10, integrating both sides once will give us f'(x):
∫ f''(x) dx = ∫ 10 dx

Integrating 10 with respect to x will give us 10x + C1, where C1 is the constant of integration.

Therefore, f'(x) = 10x + C1.

Step 2: Find f(x)
Now, we need to find the antiderivative of f'(x) to get f(x). Integrate f'(x) = 10x + C1 with respect to x:

∫(10x + C1) dx = ∫10x dx + ∫C1 dx

Using the power rule for integration (integration of x^n is (1/n+1) * x^(n+1)), we integrate both terms separately:

∫10x dx = 10/2 * x^2 = 5x^2 + C2, where C2 is another constant of integration.

∫C1 dx = C1x + C3, where C3 is yet another constant of integration.

Now, we have f(x) = 5x^2 + C2 + C1x + C3.

Step 3: Apply initial conditions
We need to use the given initial conditions f'(4) = 0 and f(2) = 4 to find the exact values of the constants C1, C2, and C3.

From f'(x) = 10x + C1, we know that f'(4) = 10(4) + C1 = 40 + C1. Since f'(4) = 0, we can solve for C1:

40 + C1 = 0
C1 = -40

Now, we have f'(x) = 10x - 40.

Next, we can use the initial condition f(2) = 4 to find the value of C2 and C3. From f(x) = 5x^2 + C2 + C1x + C3, we can substitute x = 2 and f(2) = 4:

5(2)^2 + C2 + C1(2) + C3 = 4
20 + C2 - 40 + C3 = 4
C2 + C3 = 24

Since C1 = -40, we substitute -40 for C1 in C2 + C3 = 24:

C2 + C3 = 24
C2 - 40 + C3 = 24
C2 + C3 = 64

Finally, we have the values of C1, C2, and C3. Therefore, the function f(x) is:

f(x) = 5x^2 - 40x + 64