High School

Question 3:

Let \( f(x) = 2 - 1 \). Find the interval \((a, b)\) where \( y \) increases. As your answer, please input \( a + b \).

Question 4:

Let \( f(x) = x^5 - 6x^3 - 60x^2 + 5x + 3 \). Find all solutions to the equation \( f'(x) = 0 \). As your answer, please enter the sum of values of \( x \) for which \( f'(x) = 0 \).

Answer :

The interval where y increases for the function f(x) = (4x² - 1)/(x² + 1) is (-∞, -0.5) U (0.5, ∞) is 0.5-(-∞) = ∞.

To find the intervals where the function f(x) = (4x² - 1)/(x² + 1) increases, we need to find its derivative and determine its sign. The derivative of f(x) can be found using the quotient rule:

f'(x) = [(8x)(x² + 1) - (4x² - 1)(2x)]/(x² + 1)²

Simplifying this expression, we get:

f'(x) = (12x - 4x³)/(x² + 1)²

To find the critical points, we need to solve the equation f'(x) = 0:

12x - 4x³ = 0

4x(3 - x²) = 0

This gives us the critical points x = 0 and x = ±√3. We can now test the intervals between these critical points to determine the sign of f'(x) in each interval.

Testing x < -√3, we choose x = -4, and we get f'(-4) = (-224)/(17²) < 0. Therefore, f(x) is decreasing on this interval.

Testing -√3 < x < 0, we choose x = -1, and we get f'(-1) = (16)/(2²) > 0. Therefore, f(x) is increasing on this interval.

Testing 0 < x < √3, we choose x = 1, and we get f'(1) = (16)/(2²) > 0. Therefore, f(x) is increasing on this interval.

Testing x > √3, we choose x = 4, and we get f'(4) = (-224)/(17²) < 0. Therefore, f(x) is decreasing on this interval.

Hence, the interval where f(x) increases is (-∞, -0.5) U (0.5, ∞). Therefore, the answer is 0.5 - (-∞) = ∞.

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