drawing functions

Question

I need to draw a function that increases at (-2,3) and decreases at (negative infinity, -2) or (3, infinity).

Answer 1

So it has to BOUNCE at -2, which means x=-2

is a solution of multiplicity 2

The relative max occurs at x=3, which means the

derivative is zero there

With that being said, there will be another solution

GREATER than 3.

So we got y = f(x) = -(x+2)^2 (x-k), k>3

and

f'(3) = 0

f(x) = -[(x^2 + 4x + 4)(x-k)]

= -[x^3 + 4x^2 + 4x - kx^2 - 4k x - 4k]

= -[x^3 + (4-k)x^2 + (4 - 4k) x - 4k]

Then

f'(x) = -[3x^2 + 2(4-k) x + (4-4k)]

f'(3) = 0 --> 27 + 2(4-k)*3 + (4-4k) = 0

27 + 24 - 6k + 4 - 4k = 0

-10k + 55 = 0

k = -55/-10 = 5.5

The function is f(x) = -x^3 + 1.5 x^2 +18x +22