For the function: $$ f(x)=e^{x}-x^{2} $$ find the inflection point. $$ \qquad $$ (In [?], [ $$ ]) $$

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1. Compute the first derivative: $$ f'(x)=\frac{d}{dx}\left(e^x-x^2\right)=e^x-2x $$ 2. Compute the second derivative: $$ f''(x)=\frac{d}{dx}\left(e^x-2x\right)=e^x-2 $$ 3. Set the second derivative equal to zero to find the potential inflection point: $$ e^x-2=0 \quad \Rightarrow \quad e^x=2 \quad \Rightarrow \quad x=\ln 2 $$ 4. To confirm that the concavity changes at $$ x=\ln 2 $$, observe that: - For $$ x < \ln 2 $$: $$ e^x < 2 $$ so $$ f''(x) < 0 $$ (concave down). - For $$ x > \ln 2 $$: $$ e^x > 2 $$ so $$ f''(x) > 0 $$ (concave up). 5. Find the corresponding $$ y $$-coordinate by evaluating $$ f(x) $$ at $$ x=\ln 2 $$: $$ f(\ln 2)=e^{\ln 2}-(\ln 2)^2=2-(\ln 2)^2 $$ The inflection point is: $$ \left(\ln 2,\; 2-(\ln 2)^2\right) $$ The inflection point of the function $$ f(x) = e^{x} - x^{2} $$ is at $$ \left(\ln 2,\; 2 - (\ln 2)^2\right) $$.

For the function:

find the inflection point.
(In [?], [

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For the function: find the inflection point. (In [?], [