Calculate m and n of function g(x) knowing it admits an local minimum in point (2;4)

Question

Let $g(x) = 3x^2 + mx + n$

I'm trying to solve this problem.

• First, I calculate $g(2) = 4$, which implies $-8 = 2m + n$
• After, I calculate the derivative of $g(x)$, which is $g'(x) = 6x + m$

The problem I'm facing right now is that I need the exact value of $g'(x)$ to get $m$.

What can I do to solve this problem?

Answer 1

Quadratic with coefficient of $x^2$ +ve is upward opening parabola with minimum at vertex. $$Let \,\,y=ax^2+bx+c $$ Minimum exist at vertex whose coordinates are $(-\frac{b}{2a},-\frac{D}{4a})$ $$ $$ For given equation a=3 , b=m , c=n $$\therefore -\frac{m}{6}=2 \,\,and\,\,-(\frac{m^2-12n}{12})=4$$ $$\therefore m=-12 , n=16 $$

Answer 2

We have $ g'(2)=0$. Can you proceed?

Answer 3

You are correct. But since $(2,4)$ is a local minimum of $g$, we must have $g'(2)=0$, so you can now complete the problem.

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An alternative approach is from analytic geometry. Note that $g(x)$ is a parabola, so its minimum must be its vertex, which for a parabola $ax^2+bx+c$ always lies at $x = -\frac{b}{2a}$, which in your case yields $2 = -\frac{m}{6}$, which makes the problem equivalently trivial...

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First one may be a little quicker, but the second one requires no calculus whatsoever.