Find a cubic function f(x)=ax^3+cx^2+d that has a local maximum value of 9 at -4 and a local minimum value of 6 at 0. Find a, c, and d.

Question

I know that d = 6 but I am worked for hours trying to find a and c. Please help!!

Answer 1

f(x) = ax^3 + cx^2 + d
f'(x) = 3ax^2 + 2cx + 0

f'(-4) = 9
48a - 8c = 9 **

the point (-4,9) is on f(x)
-64a + 16c + 6 = 9 ***

2 times ** ---> 96a - 16c = 18
*** ---------> -64a + 16c = 3
add them:
32a = 21
a = 21/32

in 48a - 8c = 9
48(21/32) - 8c = 9
-8c = -45/2
c = 45/16

check my arithmetic, expected "nicer" numbers

Answer 2

check the related questions below.

for extrema as described above, you need

f' = 3ax^2+2cx = 0 at x=0 and -4, so
x(3ax+2c) = 0
-12a+2c = 0
You also have two points, so
-64a+16c+d = 9
0a+0c+d = 6

a = 3/32
c = 9/16
d = 6

so, that means that

f(x) = 3/32 x^3 + 9/16 x^2 + 6

the url below confirms the conditions we wanted.

https://www.wolframalpha.com/input/?i=3%2F32+x^3+%2B+9%2F16+x^2+%2B+6