Q:
If $b>0$ then find the number of values of'a' for which domain and range of $f\left(x\right)=\sqrt{ax^{2}+bx}$ are equal.
My approach:
Since we have a square root, $$ax^{2}+bx\ge0\to x\left(ax+b\right)\ge0$$
If $a>0$ then domain is: $$x \in \ \left(-\infty,-\frac{b}{a}\right] \cup \ \left[0,\infty\right)$$
If $a<0$ then domain is:
$$[0,-\frac{b}{a}]$$
If $a=0$ then domain is: $$x\ge0$$
But I don't know how to proceed further with the main question. Any hints or suggestions are welcome.
Answer:
2 values of a are possible.
Best Answer
If $a < 0$ the domain you pointed out is wrong , the right one is $[0,-\frac{b}{a}]$.
One value is $a = 0$ because then $f(x) = \sqrt{bx}$ which has range $[0,\infty]$.
For $a > 0$ there can't be no value having domain = range because your domain has always negative values inside while range, because of the presence of the square root, needs to contain just positive values.
So the second value has $a < 0$, now you can see that $f(0) = f(-\frac{b}{a}) = 0$ and that is the minimum value of the range , consequently the maximum is attained inside the interval so you can just use a derivative test to find which one is it :
$$f'(x) = 0 \implies \frac{2ax+b}{2 \sqrt{ax^2+bx}} = 0 \implies x = -\frac{b}{2a}$$
and so the maximum is $f(-\frac{b}{2a}) = b\sqrt{\frac{1}{4a}-\frac{1}{2a}}$, but then to have domain = range you need that
$$b\sqrt{\frac{1}{4a}-\frac{1}{2a}} = -\frac{b}{a} \implies a =-4$$