Question

Find the local minima if f(x)=e^x(1-x)^2

Answer 1

0

Answer 2

To find the local minima of 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 $v = (1-x)^2$: $u' = e^x$ $v' = 2(1-x)(-1) = -2(1-x)$ So, $f'(x) = e^x(1-x)^2 + e^x(-2(1-x))$ Factor out $e^x(1-x)$: $f'(x) = e^x(1-x)[(1-x) - 2]$ $f'(x) = e^x(1-x)(-1-x)$ $f'(x) = -e^x(1-x)(1+x)$ $f'(x) = -e^x(1-x^2)$

2. Find the critical points by setting $f'(x) = 0$: $-e^x(1-x^2) = 0$ Since $e^x > 0$ for all real $x$, we must have: $1-x^2 = 0$ $x^2 = 1$ $x = \pm 1$ The critical points are $x = 1$ and $x = -1$.

3. Find the second derivative, $f''(x)$: Using the product rule on $f'(x) = -e^x(1-x^2)$, where $u = -e^x$ and $v = (1-x^2)$: $u' = -e^x$ $v' = -2x$ So, $f''(x) = (-e^x)(1-x^2) + (-e^x)(-2x)$ $f''(x) = -e^x(1-x^2) + 2xe^x$ Factor out $e^x$: $f''(x) = e^x(-1+x^2+2x)$ $f''(x) = e^x(x^2+2x-1)$

4. Apply the second derivative test:

   • At $x = 1$: Substitute $x=1$ into $f''(x)$: $f''(1) = e^1(1^2+2(1)-1)$ $f''(1) = e(1+2-1)$ $f''(1) = 2e$ Since $f''(1) = 2e > 0$, $x=1$ is a point of local minimum.

   • At $x = -1$: Substitute $x=-1$ into $f''(x)$: $f''(-1) = e^{-1}((-1)^2+2(-1)-1)$ $f''(-1) = e^{-1}(1-2-1)$ $f''(-1) = -2e^{-1}$ Since $f''(-1) = -2e^{-1} < 0$, $x=-1$ is a point of local maximum.

5. Calculate the local minimum value: The local minimum occurs at $x=1$. Substitute this value back into the original function $f(x)=e^x(1-x)^2$: $f(1) = e^1(1-1)^2$ $f(1) = e(0)^2$ $f(1) = 0$

The local minimum value of the function is $0$, which occurs at $x=1$.