What are the absolute extrema of # f(x)= xe^(x^2)/128in [-5,16]#?

Question

Charles Anctil · Accepted Answer

#f(16)=e^256/8 and f(0)=0# #f'=(1/128)(xe^(x^2))'=(1/128)((x)'e^(x^2)+(e^(x^2))'x)=(1/128)e^(x^2)(1+2x^2)# Here, #e^(x^2)>=1 and 1+2x^2>=1#. So, f'>=1>0# And so, f is an increasing function in #x in (-oo, oo)# As #|f(-5)|=(5/128)e^25 < f(16) = e^256/8#, the absolute maximum =#f(16)=e^256/8# and, as #|f| >= 0 and f(0) = 0#, the absolute minimum=#f(0)=0#.

Emily Ammerman · Answer

undefined To find the absolute extrema of the function $f(x) = \frac{x e^{x^2}}{128}$ on the interval $[-5, 16]$, we need to follow these steps:

1. **Find the derivative of $f(x)$** to locate critical points inside $[-5, 16]$.
2. **Evaluate $f(x)$ at critical points and the endpoints of the interval** to identify the absolute maximum and minimum.

### 1. Find the derivative of $f(x)$

$$f(x) = \frac{x e^{x^2}}{128}$$

First, apply the product rule $d(uv) = u dv + v du$ and the chain rule $d(e^u) = e^u du$:

$$f'(x) = \frac{d}{dx}\left(x\right) \cdot \frac{e^{x^2}}{128} + x \cdot \frac{d}{dx}\left(\frac{e^{x^2}}{128}\right)$$

$$f'(x) = \frac{e^{x^2}}{128} + x \cdot \left(\frac{2x e^{x^2}}{128}\right)$$

$$f'(x) = \frac{e^{x^2}}{128} + \frac{2x^2 e^{x^2}}{128}$$

$$f'(x) = \frac{e^{x^2}(1 + 2x^2)}{128}$$

### 2. Solve $f'(x) = 0$ to find critical points

Set the derivative equal to zero and solve for $x$:

$$\frac{e^{x^2}(1 + 2x^2)}{128} = 0$$

Since $e^{x^2} > 0$ for all real numbers $x$, the equation simplifies to finding $x$ such that:

$$1 + 2x^2 = 0$$

$$2x^2 = -1$$

This equation has no real solutions because $x^2$ cannot be negative. Thus, there are no critical points in the real number system within or outside the interval.

### 3. Evaluate $f(x)$ at the interval's endpoints

Since there are no critical points within $[-5, 16]$, the absolute extrema occur at the endpoints of this interval. Evaluate $f(x)$ at $x = -5$ and $x = 16$:

$$f(-5) = \frac{-5 e^{(-5)^2}}{128} = \frac{-5 e^{25}}{128}$$

$$f(16) = \frac{16 e^{(16)^2}}{128} = \frac{16 e^{256}}{128}$$

### Conclusion

The absolute minimum and maximum of $f(x) = \frac{x e^{x^2}}{128}$ on the interval $[-5, 16]$ must be at the endpoints since there are no critical points in the interval. To determine which is the absolute max or min, consider the values:

- At $x = -5$, $f(x)$ takes on a relatively small (negative) value due to the negative exponent in $e^{25}$.
- At $x = 16$, $f(x)$ takes on a large positive value due to the exponent $e^{256}$, which significantly exceeds the negative value at $x = -5$.

Therefore, the absolute minimum is at $x = -5$, and the absolute maximum is at $x = 16$, with their corresponding $f(x)$ values being the min and max values, respectively.