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. Find a, c, and d.
I know that d = 6 but I am worked for hours trying to find a and c. Please help!!
f(x) = ax^3 + cx^2 + d
f'(x) = 3ax^2 + 2cx + 0
f'(-4) = 9
48a - 8c = 9 **
the point (-4,9) is on f(x)
-64a + 16c + 6 = 9 ***
2 times ** ---> 96a - 16c = 18
*** ---------> -64a + 16c = 3
add them:
32a = 21
a = 21/32
in 48a - 8c = 9
48(21/32) - 8c = 9
-8c = -45/2
c = 45/16
check my arithmetic, expected "nicer" numbers
check the related questions below.
for extrema as described above, you need
f' = 3ax^2+2cx = 0 at x=0 and -4, so
x(3ax+2c) = 0
-12a+2c = 0
You also have two points, so
-64a+16c+d = 9
0a+0c+d = 6
a = 3/32
c = 9/16
d = 6
so, that means that
f(x) = 3/32 x^3 + 9/16 x^2 + 6
the url below confirms the conditions we wanted.
https://www.wolframalpha.com/input/?i=3%2F32+x^3+%2B+9%2F16+x^2+%2B+6
To find the values of a and c, we can use the information about the local maximum and minimum.
First, let's consider the local maximum value of 9 at x = -4. This means that the point (-4, 9) is on the graph of the cubic function. Plugging in these values, we get:
9 = a(-4)^3 + c(-4)^2 + d
9 = -64a + 16c + 6 (1)
Next, let's consider the local minimum value of 6 at x = 0. This means that the point (0, 6) is on the graph of the cubic function. Plugging in these values, we get:
6 = a(0)^3 + c(0)^2 + d
6 = 0a + 0c + 6
6 = 6 (2)
From equation (2), we can see that d = 6.
Now we can substitute this value of d into equation (1):
9 = -64a + 16c + 6
Rearranging this equation, we get:
-64a + 16c = 3
Next, we want to express one variable in terms of the other. Let's solve equation (2) for c:
c = (3 + 64a) / 16
Now we can substitute this value of c into equation (1):
-64a + 16((3 + 64a)/16) = 3
-64a + 3 + 64a = 3
3 = 3
We can see that this equation is true, which means that any value of a and c that satisfy c = (3 + 64a) / 16 will work. Therefore, the values of a and c can be any real numbers.
To summarize:
- We know that d = 6.
- The values of a and c can be any real numbers.
So, we cannot determine unique values for a and c based on the given information.