FIND F. f"(x) = 24x^2 + 2x + 10, f(1) = 5 and f ' (1) = -3.

I got up to here 2x^4 + (x^3)/3 + 5x^2 + CX + D.

someone show me how to find C and D with steps please.

f "(x) = 24x² + 2x + 10

∫ f "(x) dx = f '(x)
∫ (24x² + 2x + 10) dx = f '(x)
f '(x) = 24/3 x³ + 2/2 x² + 10x + C
f '(x) = 8x³ + x² + 10x + C

Given that f '(1) = -3,
-3 = 8(1)³ + (1)² + 10(1) + C
-3 = 8 + 1 + 10 + C
C = -22

f '(x) = 8x³ + x² + 10x - 22
∫ f '(x) dx = f(x)
∫ (8x³ + x² + 10x - 22) dx = f(x)
f(x) = 8/4 x⁴ + 1/3 x³ + 10/2 x² - 22x + D
f(x) = 2x⁴ + 1/3 x³ + 5x² - 22x + D

Given that f(1) = 5,
5 = 2(1)⁴ + 1/3 (1)³ + 5(1)² - 22(1) + D
5 = 2 + 1/3 + 5 - 22 + D
D = 59/3

f(x) = 2x⁴ + 1/3 x³ + 5x² - 22x + 59/3

f '(x) = 8x^3 + x^2 + 10x + C

Plugging in initial condition:

-3 = 8 + 1 + 10 + C --> C = -22

f '(x) = 8x^3 + x^2 + 10x - 22

f(x) = 2x^4 + (x^3)/3 + 5x^2 - 22x + D

5 = 2 + 1/3 + 5 - 22 + D

5 = 1/3 - 15 + D

5 = -44/3 + D

D = 59/3

f(x) = 2x^4 + (x^3)/3 + 5x^2 - 22x + 59/3