anonymous
  • anonymous
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 ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[f(x)=\sqrt{c-x^{2}} \text{, set the inside equal to 0 and solve.}\]
anonymous
  • anonymous
after that, set x = -5 and 5 to get c.
anonymous
  • anonymous
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 ?

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anonymous
  • anonymous
yes, sorry. that's probably better.
anonymous
  • anonymous
\[c-x^{2} \ge 0\]
anonymous
  • anonymous
mult by -1 to get \[x^{2}-c \le 0\]
anonymous
  • anonymous
its no problem. i just don't know what to do next ? (c+sqrt(x))(c-sqrt(x))<=0
anonymous
  • anonymous
so now move c over... \[x^{2} \le c\]
anonymous
  • anonymous
what should we do next?
anonymous
  • anonymous
plug in -5 and 5
anonymous
  • anonymous
not yet, we need to solve for x.
DebbieG
  • DebbieG
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}
anonymous
  • anonymous
\[-\sqrt{c} \le x \le \sqrt{c} \text{, set } -\sqrt{c}=-5 \text{ and set } \sqrt{c}=5\]
anonymous
  • anonymous
ohh got it. Thank you sooo much :)

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