Find a cubic function f(x)=ax^(3)+cx^(2)+d that has a local maximum value of 9 at -4 and a local minimum value of 6 at 0.

Question

a = ?

b = ?

c = ?

Answer 1

Given:

x = 0, y = 6 so d = 6

x = -4, y = 9

find f'

f' = 3ax² + 2cx = x(3ax + 2c)

set f' = 0 to find min/max

3ax + 2c = 0 with x = -4

3a.(-4) + 2c = 0

-12a + 2c = 0 so a = c

since x = -4, y = 9

So f(-4) = a(-4)³ + c(-4)² + 6 = 9

since a = c

-64a + 16a = 3

a = -1/16

c = -1/16

Answer 2

Hello Katie.

Some of the answers needed come from the original function, others will come from the derivative. We know that (-4,9) and (0,6) are on the graph of f(x).

Since f(0)=6, f(0)=a(0)³ + c(0)² + d = 6

= 0 + 0 + d = 6, so d=6.

Now our function is

f(x)=ax³ + bx² +6

f(-4) = a(-4)³ + b(-4)² + 6 = 9

Subtract 6 from both sides

-64a + 16 b = 3

Using the derivative, f '(x) = (3a)x² + (2c)x

Minimums and maximums occur at the x's that make f '(x)=0, so we will set our f '(x)=0

0 = (3a)x² + (2c)x

Factor out an x

0 = x[ (3a)x + (2c) ]

So x = 0 there will be a min or max, but the problem states that it is a minimum at (0,6).

The other critical number comes from (3a)x + 2c = 0

Subtract 2c from both sides

(3a)x = -2c

We were told that this maximum occurs at x = -4, so now we plug in -4 for x.

-12a = -2c

Divide both sides by -2, and get

6a = c

Using the expression we got above from f(-4)

-64a + 16c = 3, we will substitute "6a" for the "c".

-64a + 16(6a) = 3

-64a + 96a = 3

32a = 3

a = 3/32

c = 6a = 9/16

Final function is:

f(x)=(3/32)x³ + (9/16)x² + 6

a = 3/32

c = 9/16

d = 6