Finding value(s) for x where the tangent line of f(x) has a slope of 4.

Question

I have this equation $\frac{3}{4}x^4 - 4x^3 + \frac{1}{2}x^2 + 7$

I am trying to find the value(s) for $x$ where the tangent line of $f(x)$ has a slope of 4.

So I take my first derivative which I get $3x^3 - 12x^2 + x$. With this equation do I want to set it to 4 and try factoring my derivative and setting them to 4?

Not really sure if that is the proper set up. Thanks!

EDIT : Edited derivative as I had typed it wrong.

Answer 1

Yes, but you made a typo in the derivative.

$$ 3x^3 - 12 x^2 + x = 4\\ 3x^3 - 12 x^2 + x - 4 = 0\\ (3x^3 + x) - (12 x^2 + 4) = 0\\ x(3x^2 + 1) - 4(3x^2 + 1) = 0\\ (x-4)(3x^2+1)=0 $$

so $x=4$ is the only real solution.

Answer 2

You should solve $f'(x) = 4$ for $x$. You'll find one real solution ($x=4$) and two imaginary solutions.