Question

Suppose f(x) is a continuous function for every value of x whose first derivative is f'(x) = 2(1 - x^2)/(1 + x^2) and f''(x) = 4x(x^2 - 3)/(1 + x^2)^2. Further, assume that it is known that f has a horizontal asymptote at y = 0. Critical values? What intervals is f(x) increasing? What intervals is f(x) decreasing? For which value(s) of x does f(x) have a local minimum? For which value(s) of x does f(x) have a local maximum? Inflection points What interval(s) does f(x) concave up? What interval(s) does f(x) concave down?

          Suppose f(x) is a continuous function for every value of x whose first derivative is f'(x) = 2(1 - x^2)/(1 + x^2) and f''(x) = 4x(x^2 - 3)/(1 + x^2)^2. Further, assume that it is known that f has a horizontal asymptote at y = 0.
Critical values?
What intervals is f(x) increasing?
What intervals is f(x) decreasing?
For which value(s) of x does f(x) have a local minimum?
For which value(s) of x does f(x) have a local maximum?
Inflection points
What interval(s) does f(x) concave up?
What interval(s) does f(x) concave down?

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Added by Matthew W.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Vincenzo Zaccaro

06/01/2023

Step 1

Critical values: To find the critical values, we need to find the values of x for which f'(x) = 0 or f'(x) is undefined. In this case, f'(x) is defined for all x, so we just need to find where f'(x) = 0. f'(x) = 2(1 - x^2)/(1 + x^2) = 0 1 - x^2 = 0 x^2 = 1 x = Show more…

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Suppose f(x) is a continuous function for every value of x whose first derivative is f'(x) = 2(1 - x^2)/(1 + x^2) and f''(x) = 4x(x^2 - 3)/(1 + x^2)^2. Further, assume that it is known that f has a horizontal asymptote at y = 0. Critical values? What intervals is f(x) increasing? What intervals is f(x) decreasing? For which value(s) of x does f(x) have a local minimum? For which value(s) of x does f(x) have a local maximum? Inflection points What interval(s) does f(x) concave up? What interval(s) does f(x) concave down?

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