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Find the number of real roots of the equation $2^x = 1+x^2$

My try: Let we take $f(x) = 2^x-1-x^2$. Now for Drawing Graph of given function, we use Derivative Test.

$f'(x) = 2^x \cdot \ln (2)-2x$ and $f''(x) = 2^x \ln (2)-2$ and $f'''(x) = 2^x \ln(2)>0\;\forall x\in \mathbb{R}$

Means $f''(x)$ is an Strictly Increasing function. Or $f''(x)$ is an Concave upward function.

Now I did not understand how can I calculate nature of $f'(x)$ using higher derivatives. Please explain, thanks.

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  • $\begingroup$ Your $f''$ and $f'''$ are wrong. Note $(2^x)'=2^x \log 2$. $\endgroup$
    – ftfish
    Sep 27 '13 at 10:33
  • $\begingroup$ It's an interesting equation, because it has 0 and 1 as obvious roots $\endgroup$
    – Yuriy S
    Feb 4 '16 at 17:32
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Consider $$f(x)=\frac{2^x}{1+x^2}-1.$$ You will easily calculate the intervals where $f$ is increasing and decreasing. Now $f(0)=0$, you will find a max of positive value, then a min of negative value. Then have in mind that $f(5)>0$, e.g.

So there are exactly three solutions on the reel line.

Michael

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I believe you meant to say "Or $f(x)$ is a Convcave upward function", instead of $f''(x)$.

If a function is concave upwards, then both of it's endpoints go up +infinity, and it has exactly 1 minimum.

How do can the x axis intersect something like this? If the minimum is below the X axis, then twice. If the minimum is above the X axis, then zero times.

You can use the derivative to estimate where the minimum is (don't try to actually solve it, you can't without some advanced functions). Then check the value at the minimum.

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Given that the question is only to find the no. of roots, let me suggest an alternate solution. Plot graphs with your hand only for a few integral points. Now, you can find where the graphs intersect. These will give you the no. of solutions. This is much better than the calculus approach for problems which involve functions whose graphs can easily be plotted here's my rough plot

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