Determine the constants a, b, c and d so that the curve defined by y = ax^3 + bx^2 +cx +dy=ax 3 +bx 2 +cx+d has a local maximum at the point (1, -7)(2,4) and a point of inflection at (2,-11)

Answer 1

here is one very similar to what you ask. Follow its method, and post your work if you get stuck.

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