find the value of c so that the natural domain of the following function is [-5,5]
f(x)=sqrt(c-x^2)

I know the answer is c=25 but how do i prove it ?

Question

find the value of c so that the natural domain of the following function is [-5,5]
f(x)=sqrt(c-x^2)

I know the answer is c=25 but how do i prove it ?

anonymous · Answer

$$f(x)=\sqrt{c-x^{2}} \text{, set the inside equal to 0 and solve.}$$

anonymous · Answer

after that, set x = -5 and 5 to get c.

anonymous · Answer

why are we setting it equal to zero?
somebody told me to set the f(x)>0 and solve the inequality ? would this work too ?

anonymous · Answer

yes, sorry.  that's probably better.

anonymous · Answer

$$c-x^{2} \ge 0$$

anonymous · Answer

mult by -1 to get

$$x^{2}-c \le 0$$

anonymous · Answer

its no problem. i just don't know what to do next ?
(c+sqrt(x))(c-sqrt(x))<=0

anonymous · Answer

so now move c over...

$$x^{2} \le c$$

anonymous · Answer

what should we do next?

anonymous · Answer

plug in -5 and 5

anonymous · Answer

not yet, we need to solve for x.

debbieg · Answer

If you just want to PROVE that c=25 works, just find the domain of $y=\sqrt{25-x^2}$... you'll get the required domain given above.

If you want to know how to FIND c, that's a little bit different.  But notice that:

For  $y={c-x^2}$ you can say that $y=({\sqrt{c}-x})({\sqrt{c}+x})$ which is really just a downward-opening parabola with roots at $\pm \sqrt{c}$, right?

so the domain of the square root function will be $ -\sqrt{c}<x<\sqrt{c}$

so just square either endpoint of the given domain to find c.

anonymous · Answer

$$-\sqrt{c} \le x \le \sqrt{c} \text{, set } -\sqrt{c}=-5 \text{ and set } \sqrt{c}=5$$

anonymous · Answer

ohh got it. 
Thank you sooo much :)