Question

If f'(x) = 4x³ − 3x²  + 2x + k, f(0) = 1 and f(1) = 4, find f(x).

Answer 1

f'(x) = 4x³ − 3x² + 2x + k  ....[Given]

f(x) = ∫ f'(x) dx

= ∫ (4x³ − 3x² + 2x + k) dx

= 4 ∫ x³ dx − 3 ∫ x² dx + 2 ∫ x dx + k ∫ dx

= `4 ("x"^4/4) - 3("x"^3/3) + 2("x"^2/2)  "kx" + "c"`

∴ f(x) = x⁴ − x³ + x² + kx + c  ....(i)

Now, f(0) = 1  ...[Given]

∴ (0)⁴ − (0)³ + (0)² + k(0) + c = 1

∴ c = 1  ....(ii)

Also, f(1) = 4  ....[Given]

∴ 1 − 1 + 1 + k + c = 4

∴ 1 + k + 1 = 4

∴ 2 + k = 4

∴ k = 2  ...(iii)

Substituting (ii) and (iii) in (i), we get

f(x) = x⁴ − x³ + x² + 2x + 1