The local extrema correspond to zeroes of the derivative. That derivative is a wauadratic function. So, you should write down a quadratic function that has its zeroes at x = -2 and x = 1. Then integrate that function. This function is fixed up to an overall factor.
Then you integrate this function and fix the overall factor and the integration constant by demanding that the correct values are assumed at the local extrema.
Find the cubic function f(x)=ax^3+bx^2+cx+d that has a local max value of 3 at -2 and a local min values of 0 at 1.
I have no idea how to solve this. Could you please give step by step answers and explanations for the steps. Thanks!
6 answers
so
f(1)=a+b+c+d=0
f(-2)=-8a+4b-2c+d=3
f'(1)=3a+2b+c=0
f'(-2)=12a-4b+c=0
Now do I solve the system of equations for a,b,c,d? But is it possible to solve for this system of equations with 4 unknowns? Can I only substitute derivative into derivative and original into original or does it not matter?
f(1)=a+b+c+d=0
f(-2)=-8a+4b-2c+d=3
f'(1)=3a+2b+c=0
f'(-2)=12a-4b+c=0
Now do I solve the system of equations for a,b,c,d? But is it possible to solve for this system of equations with 4 unknowns? Can I only substitute derivative into derivative and original into original or does it not matter?
you can solve them anyway you want. I would use matrix algebra.
I would suggest you start with this:
f'(x) = C(x+2)(x-1)
Then what is the general form of f(x)?
f'(x) = C(x+2)(x-1)
Then what is the general form of f(x)?
Could you explain how to solve? How would you do matrix algebra here?
f'(x) = C(x+2)(x-1) =
C(x^2 + x - 2)
f(x) = C(1/3 x^3 + 1/2 x^2 - 2 x) + p
We need to satisfy the equations:
f(-2) = 3
f(1) = 0
So, we have:
10/3 C + p = 3
-7/6 C + p = 0
The solution is:
C = 2/3 and p = 7/9
C(x^2 + x - 2)
f(x) = C(1/3 x^3 + 1/2 x^2 - 2 x) + p
We need to satisfy the equations:
f(-2) = 3
f(1) = 0
So, we have:
10/3 C + p = 3
-7/6 C + p = 0
The solution is:
C = 2/3 and p = 7/9
2 answers