clearly, we want
f'(x) = 3ax^2 + 2bx + c
to be zero at x= -5 and 4. So
3ax^2 + 2bx + c = k(x+5)(x-4) = kx^2+kx-20k
so, c=-20k, 2b=k, 3a=k and
f(x) = (k/3)x^3 + (k/2)x^2 - 20kx + d
f(4) = 64k/3 + 8k - 80k + d = -152k/3 + d = -4
f(-5) = -125k/3 + 25k/2 + 100k + d = 425k/6 + d = 4
solve those two and you get
k = 48/729
d = -484/729
f(x) = 16/729 x^3 + 24/729 x^2 - 960/729 x + -484/729
or,
1/729 (16x^3 + 24x^2 - 960x - 484)
You can see the curve at
http://www.wolframalpha.com/input/?i=16%2F729+x^3+%2B+24%2F729+x^2+-+960%2F729+x+%2B+-484%2F729
scroll down to where it lists the relative max,min to verify that it meets the requirements.
Find a cubic polynomial such as f(x)=ax^3+bx^2+cx+d
that has horizontal tangents at the points (-5,4) and (4,-4)
1 answer