Calculate the number of real numbers $k$ such that $f(k) = 2$ if $f(x) = x^4 − 3x^3 − 9x^2 + 4$. 
Calculate the number of real numbers $k$ such that $f(k) = 2$ if $f(x) = x^4 − 3x^3 − 9x^2 + 4$.

How do I calculate the number of values for $k$? I can't seem to understand.  Please help me solve this problem.
 A: This is to find the number of real roots of $f(x)-2$.
$$\frac{d}{dx}(f(x)-2)=4x^3-9x^2-18x$$
which is equal to $0$ when $x=0$ or $\displaystyle \frac{9\pm4\sqrt{41}}{8}$.
$$f\left(\frac{9-4\sqrt{41}}{8}\right)-2=\frac{-23195-3321\sqrt{41}}{512}<0$$
$$f(0)-2=2>0$$
$$f\left(\frac{9+4\sqrt{41}}{8}\right)-2=\frac{-23195-3321\sqrt{41}}{512}<0$$
So $f(x)-2$ has a zero in each of the intervals $(-\infty,\frac{9-4\sqrt{41}}{8})$, $(\frac{9-4\sqrt{41}}{8},0)$, $(0,\frac{9+4\sqrt{41}}{8})$ and $(\frac{9+4\sqrt{41}}{8},\infty)$.
There are $4$ solutions.
You can also try the Descartes' rule of signs 
https://en.wikipedia.org/wiki/Descartes%27_rule_of_signs
A: Your question is equivalent to finding the number of real roots of the equation
$$x^4-3x^3-9x^2+4=2$$
which is equivalent to finding the number of real zeros of the function:
$$f(x) = x^4-3x^3-9x^2+2=0$$
Taking the derivative, we have:
$$f'(x) = 4x^3-9x^2-18x$$
The derivative is equal to zero at three points: $x_1 = \dfrac{9 - 3\sqrt{41}}{8}, x_2 = 0$ and $x_3 = \dfrac{9 + 3\sqrt{41}}{8}$. Also note that $f(x) \to +\infty$ for both $x \to +\infty$ and $x \to -\infty$. This means that $f(x)$ is decreasing in $(-\infty, x_1)$, then increasing in $(x_1, x_2)$, then decreasing in $(x_2, x_3)$, and increasing in $(x_3, +\infty)$.
Also note that $f(x_1) < 0, f(x_2) > 0, f(x_3) < 0$. Thus, $f(x)$ intersects the $x$-axis $4$ times, so it has exactly $4$ real zeros.
The plot of $f(x)$ for illustrating the argument above (from WolframAlpha):

