SOLUTION: find the extrema of the ff. function on the given interval, if there are any. determine the values of x at which the extrema occur. f(x)= x/x^2+2 on [-1,4].

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Question 1193572: find the extrema of the ff. function on the given interval, if there are any. determine the values of x at which the extrema occur.
f(x)= x/x^2+2 on [-1,4].
Answer by parmen(42) (Show Source):
You can put this solution on YOUR website!
I've been improving my skills at solving these simplification problems. Let's find the extrema of the function:
$$f(x)=\frac{x}{x^2+2}$$
On the interval $[-1, 4]$.
We can find the extrema of a function by finding its critical points, which are the points where the derivative is zero or undefined.
**Steps to solve:**
**1. Differentiate the function:**
$$f'(x)=\frac{2-x^2}{(x^2+2)^2}$$
**2. Set the derivative equal to zero and solve for x:**
$$f'(x)=0$$
$$\frac{2-x^2}{(x^2+2)^2}=0$$
$$2-x^2=0$$
$$x=\pm\sqrt{2}$$
**3. Evaluate the function at the critical points and endpoints of the interval:**
$$f(-1)=\frac{-1}{(-1)^2+2}=-\frac{1}{3}$$
$$f(\sqrt{2})=\frac{\sqrt{2}}{(\sqrt{2})^2+2}=\frac{\sqrt{2}}{4}$$
$$f(4)=\frac{4}{4^2+2}=\frac{2}{10}$$
**4. Compare the values of the function at the critical points and endpoints to determine the extrema:**
The maximum value of the function is $\frac{\sqrt{2}}{4}$ at $x=\sqrt{2}$.
The minimum value of the function is $-\frac{1}{3}$ at $x=-1$.
**Answer:**
The extrema of the function are:
* Maximum value: $\frac{\sqrt{2}}{4}$ at $x=\sqrt{2}$
* Minimum value: $-\frac{1}{3}$ at $x=-1$