Range of $f(x)=[x^2 - 3x]$

Question

I need to find the range of this function $$f(x)=gif (x^2 - 3x.)$$ I have no idea how to start solving this. Can anyone help?

Answer 1

Complete the square: $x^2-3x=\left(x-\dfrac 32\right)^2-\dfrac 94$.

Then $f(x)=\lfloor\left(x-\dfrac 32\right)^2-\dfrac 94\rfloor$.

So the range is $\left[-3,\infty\right)\cap \Bbb Z$.

Answer 2

If we assume that $x$ can have any real value, then

$y = x^2 - 3x = (x-\frac{3}{2})^2 - \frac{9}{4} \geq - \frac{9}{4}$

Answer 3

Take the derivative of $f(x)$. Then we have $$\dfrac{d}{dx}f(x)=2x-3.$$ Solve for $x$ when this derivative equals $0$ and we have $x=\dfrac{3}{2}.$ Now plug this back into your equation $f(x)=x^{2}-3x$ and we have $f\big(\dfrac{3}{2}\big)={\big(\dfrac{3}{2}\big)^2}-3\big(\dfrac{3}{2}\big)=-\dfrac{9}{4}.$ Therefore your range is $y\geq-\dfrac{9}{4}.$

Answer 4

Hint:

$f(x)$ is continuous and $\lim\limits_{x\to\pm\infty} f(x)=\infty$ and $f(x)$ has a minimum when $f'(x)=2x-3=0$.