For what values of parameter m the function $g(x) - 2x^3 - 3x^2 + mx + 3 $ has an extremum of 10? An easier way to solve it

Question

My attempt to solve this problem is very tedious and I do not think it is the optimal method. $$g'(x) = 6x^2 -6x + m$$ $$g'(x) = 0 \Rightarrow x = \frac{3 \pm\sqrt{9-6m} }{6}$$Now, I need to subsitute this x into the equation $g(x) = 10$ and solve for it. However, this will be hard to compute. Is there an easier way to tackle this problem?

Answer 1

Hint: Let $x$ be a value of an extremum. Then you know $g'(x) = 6x^2-6x+m=0$ and $g(x) = 2x^3-3x^2+mx + 3 = 10$.

Multiply the first equation by $x$, we get $6x^3-6x^2+mx=0$.

Subtract this form the second to get $-4x^3+3x^2 + 3 = 10$.

There is a rational root of $-4x^3+3x^2 - 7 =0$. It is not very hard to find.

If you have that, you can divide that root out and solve the remaining quadratic.

Answer 2

hint: for any $x$ satisfying your requirement you have: $$ 2x^3-3x^2+mx +3 = 10 \\ 6x^2-6x +m = 0 $$ so you can eliminate $m$ and solve the resulting cubic to find $x$