Find the function f given that the slope of the tangent line at any point (x, f(x)) is f '(x) and that the graph of f passes through the given point.

Question

f'(x) = 4(2x-7)³ (4, 3/2)

f(x) = ????

Answer 1

If we integrate f'(x) and use our initial condition (x,f(x))=(4,3/2), then we can determine the function f(x):

f'(x) = 4(2x-7)³

∫f'(x)dx = ∫4(2x-7)³dx

Let u=2x-7 and du = 2dx so that 2du = 4dx:

f(x) = ∫2u³du

f(x) = 2u⁴/4 + C

f(x) = u⁴/2 + C

f(x) = (2x-7)⁴/2 + C

Now plug in (x,f(x))=(4,3/2) to solve for C:

3/2 = (2(4)-7)⁴/2 + C

3/2 = (8-7)⁴/2 + C

3/2 = 1/2 + C

1 = C

Thus, f(x) = (2x-7)⁴/2 + 1

Hope this helped!

Answer 2