Question

Find the values of a and b so that the function f(x) = (1/3)x^3 + ax^2 + bx will have a relative extreme at (3, 1).
The answers for this are a = -19/9 and b = 11/3. I cannot seem to get these answers.

Answer 1

\[f(x)=\frac{ 1 }{ 3 }x^3 + ax^2 + bx\]

You know that in order to find extrema, you must take the differentiate the function and sent dy/dx to 0 and find the x value.
So differentiate it:

\[f'(x)=\frac{ d }{ dx }(\frac{ 1 }{ 3 }x^3 + ax^2 + bx)\]
\(f'(x)=x^2+2ax+b\)
Now you have the extrama given to you as (3,1), so the following statement is certainly true:
\(0=(3)^2+2a(3)+b\)
And you have the extrema given as (3,1), so the following statement is also true:
\(1=\frac{ 1 }{ 3 }(3)^3 + a(3)^2 + b(3)\)

So you have two systems of equations, correct?

\((1)...1=\frac{ 1 }{ 3 }(3)^3 + a(3)^2 + b(3)\)
\((2)...0=(3)^2+2a(3)+b\)

Perfect chance to use linear elimination and find a and b!

You will get \(a\) as -19/9 and \(b\) as 11/3. Viola!