Question

Find local maxima and minima of f(x)=e^x(1+x^2)

Answer 1

No local maxima and no local minima

Answer 2

To find the local maxima and minima of the function $f(x) = e^x(1+x^2)$, we follow these steps:

1. Find the first derivative, $f'(x)$: Given $f(x) = e^x(1+x^2)$. Using the product rule, $(uv)' = u'v + uv'$, where $u = e^x$ and $v = 1+x^2$: $u' = e^x$ $v' = 2x$ $f'(x) = e^x(1+x^2) + e^x(2x)$ $f'(x) = e^x(1+x^2+2x)$ $f'(x) = e^x(x^2+2x+1)$ $f'(x) = e^x(x+1)^2$

2. Find the critical points by setting $f'(x) = 0$: $e^x(x+1)^2 = 0$ Since $e^x > 0$ for all real $x$, we must have: $(x+1)^2 = 0$ $x+1 = 0$ $x = -1$ So, there is only one critical point at $x = -1$.

3. Apply the first derivative test: To determine if $x=-1$ is a local maximum, local minimum, or neither, we examine the sign of $f'(x)$ around $x=-1$. $f'(x) = e^x(x+1)^2$.

   • For $x < -1$ (e.g., $x=-2$): $f'(-2) = e^{-2}(-2+1)^2 = e^{-2}(-1)^2 = e^{-2}(1) = \frac{1}{e^2} > 0$. This means $f(x)$ is increasing for $x < -1$.
   • For $x > -1$ (e.g., $x=0$): $f'(0) = e^{0}(0+1)^2 = 1(1)^2 = 1 > 0$. This means $f(x)$ is increasing for $x > -1$.

   Since $f'(x)$ is positive on both sides of $x=-1$, the function is strictly increasing through $x=-1$. Therefore, $x=-1$ is neither a local maximum nor a local minimum.

Conclusion: The function $f(x) = e^x(1+x^2)$ has no local maxima and no local minima.