Question

Find the function given that it satisfies f''(x)=36x^2+24x and its graph has a horizontal tangent line at the point (0,1)

Answer 1

first you must integrate F"(x) to get F'(x)
F'(x) = \[\int\limits (36x ^{2}+24x) dx = 12x ^{3}+ 12x ^{2} +c\]
because graph has horizontal tangent line at (0,1) so the value of gradient is 0
Substitute to F'(x) to find c
0 = 12(0)^3 + 12(0)^2 + c --> c = 0

Answer 2

then you integrate F'(x) to get F(x)
F(x) = \[\int\limits (12x ^{3}+12x ^{2}) dx = 3x ^{4}+4x ^{3}+ c\]
Then substitute x = 0, F(x) = 1
1 = 3(0)^4 + 4(0)^3 + c --> c = 1
Hence, F(x) = \[3x ^{4}+4x ^{3}+1\]