Question

If $\rm f'(x)=4x^{3}-3x^{2}+2x+k$, where k is constant to be determined and if f(0) = 1, f(1) = 4, find f(4).

Answer 1

Answer:

213

Step-by-step explanation:

Hi,

Given  f'(x)=4x³-3x²+2x+k

so f(x) = ∫f'(x)dx

f(x) = ∫4x³-3x²+2x+k dx

f(x) = x⁴ - x³ + x² + kx + c, where k and c are constants

Given f(0) = 1

On substituting 0 in place of x in f(x), we get

⇒ c = 1

Given f(1) = 4 ,

1 - 1 + 1 + k + 1 = 4

⇒ k = 2

Thus, f(x) = x⁴ - x³ + x² + 2x + 1.

f(4) = 4⁴ - 4³ + 4² + 8 + 1

f(4) = 217.

Hope, it helps !