Question

Find the local maxima and minima for the function f(x)=e^x(1-x^2)

Answer 1

Local maximum at x = √2-1 with value 2(√2-1)e^(√2-1), local minimum at x = -1-√2 with value -2(1+√2)e^(-1-√2)

Answer 2

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

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

2. Find the critical points by setting $f'(x) = 0$: $e^x(-x^2-2x+1) = 0$ Since $e^x > 0$ for all real $x$, we must have: $-x^2-2x+1 = 0$ $x^2+2x-1 = 0$ Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: $x = \frac{-2 \pm \sqrt{2^2-4(1)(-1)}}{2(1)}$ $x = \frac{-2 \pm \sqrt{4+4}}{2}$ $x = \frac{-2 \pm \sqrt{8}}{2}$ $x = \frac{-2 \pm 2\sqrt{2}}{2}$ $x = -1 \pm \sqrt{2}$ The critical points are $x_1 = -1+\sqrt{2}$ and $x_2 = -1-\sqrt{2}$.

3. Find the second derivative, $f''(x)$: Using the product rule on $f'(x) = e^x(-x^2-2x+1)$, where $u = e^x$ and $v = -x^2-2x+1$: $u' = e^x$ $v' = -2x-2$ $f''(x) = e^x(-x^2-2x+1) + e^x(-2x-2)$ $f''(x) = e^x(-x^2-2x+1-2x-2)$ $f''(x) = e^x(-x^2-4x-1)$ Alternatively, since $-x^2-2x+1 = 0$ at critical points, we can write $f''(x) = e^x(-2x-2)$ for evaluation at critical points.

4. Apply the second derivative test:

   • At $x_1 = -1+\sqrt{2}$: $f''(-1+\sqrt{2}) = e^{(-1+\sqrt{2})}(-2(-1+\sqrt{2})-2)$ $f''(-1+\sqrt{2}) = e^{(-1+\sqrt{2})}(2-2\sqrt{2}-2)$ $f''(-1+\sqrt{2}) = e^{(-1+\sqrt{2})}(-2\sqrt{2})$ Since $e^{(-1+\sqrt{2})} > 0$ and $-2\sqrt{2} < 0$, $f''(-1+\sqrt{2}) < 0$. Therefore, $x_1 = -1+\sqrt{2}$ is a point of local maximum.

   • At $x_2 = -1-\sqrt{2}$: $f''(-1-\sqrt{2}) = e^{(-1-\sqrt{2})}(-2(-1-\sqrt{2})-2)$ $f''(-1-\sqrt{2}) = e^{(-1-\sqrt{2})}(2+2\sqrt{2}-2)$ $f''(-1-\sqrt{2}) = e^{(-1-\sqrt{2})}(2\sqrt{2})$ Since $e^{(-1-\sqrt{2})} > 0$ and $2\sqrt{2} > 0$, $f''(-1-\sqrt{2}) > 0$. Therefore, $x_2 = -1-\sqrt{2}$ is a point of local minimum.

5. Calculate the local maximum and minimum values: For the critical points, we know that $x^2+2x-1=0$, which implies $1-x^2 = 2x$. So, $f(x) = e^x(1-x^2) = e^x(2x)$ at the critical points.

   • Local Maximum Value (at $x = -1+\sqrt{2}$): $f(-1+\sqrt{2}) = e^{(-1+\sqrt{2})}(2(-1+\sqrt{2}))$ $f(-1+\sqrt{2}) = 2(\sqrt{2}-1)e^{\sqrt{2}-1}$

   • Local Minimum Value (at $x = -1-\sqrt{2}$): $f(-1-\sqrt{2}) = e^{(-1-\sqrt{2})}(2(-1-\sqrt{2}))$ $f(-1-\sqrt{2}) = -2(1+\sqrt{2})e^{-1-\sqrt{2}}$