F''(x) = 6x + 8
∫f''(x) = f'(x) = 3x^2 + 8x + c1
∫f'(x) = f(x) = x^3 + 4x^2 + c1x + c2
At point (0, -2),
c2 = -2
3x - y = 2
y = 3x - 3
f'(0) = 3
c1 = 3
Therefore, the required function is
f(x) = x^3 + 4x^2 + 3x - 2
Answer:
The absolute value of a number is always positive. So that means you circled the correct one. GOOD JOB!!!!!!!!!!!!!!!!!!!!!!!
Step-by-step explanation:
Answer:
5)
1
Expand by distributing terms.
-2x-2\times 5−2x−2×5
2
Simplify 2\times 52×5 to 1010.
-2x-10−2x−10