Question

Maximum and minimum value

Answer 1

Answer:

Maximum: 1

Minimum: -1

Step-by-step explanation:

To solve this, we can find the derivative of the function and then solve for when the derivative is equal to 0, finding the local maximum and minimum values in this function. Then, we can compare those points to the endpoints of the interval to find the maximum and minimum.

Using the Quotient Rule, we can say that

(f/g) ' = (f'g - fg')/(g²)

$\frac{2x}{x^2+1}dx = \frac{\frac{d}{dx}(2x) * (x^2+1) - \frac{d}{dx} (x^2+1) * (2x) }{(x^2+1)^2} \\= \frac{2 *(x^2+1) - (2x)*(2x)}{(x^2+1)^2}$

Since we just need to solve for 0, we can disregard for the bottom part, resulting in

2(x²+1)-(2x)(2x) = 0

2x²+2-4x² = 0

2-2x² = 0

add 2x² to both sides to isolate the x² and its coefficient

2x² = 2

divide both sides by 2

x²=1

x = ±1

Therefore, we have 4 possible values for the maximum and minimum over this interval:

-2, -1, 0, 1

Trying each of these numbers into f(x),

f(-2) = -4/5

f(-1) = -2/2 = -1

f(0) = 0/1 = 0

f(1) = 2/2 = 1

Therefore, the maximum and minimum values of the function over the given interval are 1 and -1 respectively