Find all values of c such that f is continuous on $(-\mathrm{\infty},\mathrm{\infty}).f\left(x\right)=\{1-{x}^{2},x\le c\text{}\{x,xc$

Edmund Conti
2021-12-07
Answered

Find all values of c such that f is continuous on $(-\mathrm{\infty},\mathrm{\infty}).f\left(x\right)=\{1-{x}^{2},x\le c\text{}\{x,xc$

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Robert Harris

Answered 2021-12-08
Author has **23** answers

Since $1-{x}^{2}$ and x are continuous on $(-\mathrm{\infty},\mathrm{\infty})$ , the only possible point of discontinuity is at x=c. At this point both functions should have the same value, so

$1-{x}^{2}=x$

${x}^{2}+x-1=0$

Therefore, using quadratic formula,$x=\pm \frac{\sqrt{5}-1}{2}$

So the given function is continuous for$c=\pm \frac{\sqrt{5}-1}{2}$

Result:

$c=\pm \frac{\sqrt{5}-1}{2}$

Therefore, using quadratic formula,

So the given function is continuous for

Result:

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