# Local Extrema Question Again, With Two Parts

• Oct 26th 2009, 11:38 PM
lysserloo
Local Extrema Question Again, With Two Parts
I recently asked a question similar to this, but that question only wanted a local maximum. Now I have one asking for values to get a certain max AND min.

Problem: Find values of a, b, and c so that the function $f(x) = 2x^3 + ax^2 + bx + c$ has a local maximum at the point (-2, 22) and a local minimum at the point ( 1, -5).

So far, I've taken the derivative, which is $6x^2 + 2ax + b$, then I solved for b, which is $b = -x(6x + 2a)$. I then set b to 0 and solved for x, getting x = 0 and $x = \frac{-2a}{6}$.

Then, plugging in -2 for x, I got a = 6.

I don't think what I'm doing is right. How do I solve this taking into account both a max AND a min?
Could you please show your steps and explain a little about what you do and why? Thank you SO much! I'm really struggling with these problems.
• Oct 27th 2009, 12:04 AM
Debsta
The derivative which you have found correctly will be equal to 0 when x=-2 and x=1 (the turning points). So put -2 in for x to get an equation involving a and b.
Then do the same for x=1 to get another equation involving a and b. Solve simultaneously to find a and b.
Then put these values into f(x) so that all you still need is to find c.
Then use the fact that f(-2)=22 to find c. Check it with the other point (1, -5).