Number of real roots of $2^x = 1-x^2$ for $x\in (0,1)$

Question

How can I found no. of real roots of $2^x = 1-x^2$ in $x\in (0,1)$

I did not found a method by which i can draw graph of two curve in the interval $x\in (0,1)$

please help me , Thanks

Sorry friends actually I have mistyped the interval.

original question is

No. of real roots of $2^x = 1-x^2$ in $x\in \left(-1,0\right)$

Answer 1

Define $f(x) = 2^x + x^2 - 1$. Then $f(0) = 0$ and

$$f'(x) = 2^x \ln{2} + 2x \ge \ln{2}$$

for all $x \ge 0$. In particular, the graph of $f$ must lie above that of the line $x \ln{2}$, and we see that there are no solutions on $(0, 1)$.

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If $-1 < x < 0$, then $f''(x) = 2^{x} (\ln{2})^2 + 2$ is strictly positive, so there can be at most one solution on the interval $(-1, 0)$. On the other hand, $f(-1) = \frac{1}{2} > 0$, so there is a solution.

Answer 2

We know $2^x$ is an increasing function in $[0,1]$, and $1-x^2$ is a decreasing function in $[0,1]$. Therefore we know, the difference between left hand side and right hand side is increasing in $[0,1]$.

Also, the two sides are equal at $x=0$. This means there is no root between $(0,1)$. You can also see from the following inequality:

$$2^x>2^0 = 1-0^2> 1-x^2$$