Question

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

• A. None
• B. One
• C. Two
• D. Three
• E. Four

Answer 1

The correct option is E Four

Given, $f\left(x\right)=x^4-3x^3-9x^2+4,f\left(k\right)=2$

Therefore, $y=2=f\left(x\right)$
$\Rightarrow 2=x^4-3x^3-9x^2+4$
$\Rightarrow x^4-3x^3-9x^2+2=0$
Then plug in consecutive numbers for $x$ and if the sign of $f\left(x\right)$ changes,
$f(-2)=6$
$f(-1)=-3$ so there's one
$f\left(0\right)=2$ so that's two
$f\left(1\right)=-9$ so that's three
And eventually it has to become positive again so there are four.