f'(x) = 3ax^2 + 2bx + c = m(x+4)(x-2) = mx^2+2mx-8m
f"(x) = 6ax+2b = 2m(x+1)
since f"(-4) < 0 and f"(2) > 0, m > 0
but you already knew that, just from what you know about the shape of cubics.
Now, we have b = 3a and c = -8b and m=b, so
f(x) = b/3 x^3 + bx^2 - 8bx + d
But, we know that f(-4) = 3 and f(2) = 0, so
-64/3 b + 16b + 32b + d = 3
8/3 b + 4b - 16b + d = 0
b = 1/12, d = 7/9
which gives us
f(x) = 1/36 x^3 + 1/12 x^2 - 2/3 x + 7/9
= 1/36(x^2+3x^2-24x+28)
The desired properties can be found if you scroll down a bit at
www.wolframalpha.com/input/?i=1%2F36+x%5E3+%2B+1%2F12+x%5E2+-+2%2F3+x+%2B+7%2F9
Find a cubic function
f(x) = ax3 + bx2 + cx + d that has a local maximum value of 3 at x = −4 and a local minimum value of 0 at x = 2.
3 answers
y = a x^3 + b x^2 + c x + d
dy/dx = 3 a x^2 + b x + c is = 0 at x = -4 and at x = 2
d^2 y/dx^2 = 6 a x + b is negative at x = -4 and positive at x = 2
so at x = -4
-64 a + 16 b - 4 c+ d = 0
3 a (16) -4b +c = 0
and at x = 2
8 a +4 b + 2 c + d = 0
3 a (4)+ 2 b + c =0
find a, b and c and d
dy/dx = 3 a x^2 + b x + c is = 0 at x = -4 and at x = 2
d^2 y/dx^2 = 6 a x + b is negative at x = -4 and positive at x = 2
so at x = -4
-64 a + 16 b - 4 c+ d = 0
3 a (16) -4b +c = 0
and at x = 2
8 a +4 b + 2 c + d = 0
3 a (4)+ 2 b + c =0
find a, b and c and d
whoops, 3 not 0 at x = -4
so at x = -4
-64 a + 16 b - 4 c+ d = 3
so at x = -4
-64 a + 16 b - 4 c+ d = 3
1 answer