Determine the constants a,b,c and d so that the curve defined by the cubic
f(x) = ax3+bx2+cx+d has a local max at the point (-2,4) and a point of inflection at the point (0,0)
Patience is such a virtue ....
I just did this about 3 posts back
An inflection point is where f''(x)=0, so we find the derivative of f'(x) and plug in x=0 to find the value of b:
f''(x) = 6ax + 2b
f''(0) = 6a(0) + 2b
0 = 2b
So, since we know that b=0, we find the derivatives again:
f(x) = ax^3 + cx + d
f'(x) = 3ax^2 + c
f''(x) = 6ax
We know that at the local max, f'(-2)=0, as well as knowing that f(-2)=4 and f(0)=0. This means that d=0, and we just have "a" and "c" left:
f(-2) = a(-2)^3 + c(-2)
4 = -8a - 2c
f'(-2) = 3a(-2)^2 + c
0 = 12a + c
By solving the system of equations:
0 = 12a + c
-12a = c
4 = -8a - 2c
4 = -8a - 2(-12a)
4 = -8a + 24a
4 = 16a
4/16 = a
1/4 = a
-12a = c
-12(1/4) = c
-3 = c
Thus, the equation with local max (-2,4) and inflection point (0,0) would be f(x) = (1/4)x^3 - 3x