How do you find the local max and min for #f(x) = (3x) / (x² - 1)#?

Answer 1

#f(x) = (3x)/(x^2-1)# has no local extrema

Find the critical points by equating the first derivative to zero:

#f'(x) = frac (3(x^2-1)-6x^2) ((x^2-1)^2) = -3(x^2+1)/((x^2-1)^2)#

As the derivative is negative in all the domain of the function, the function is strictly decreasing and has no local extrema.

graph{3x/(x^2-1) [-10, 10, -5, 5]}

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Answer 2

Aurora Alvarado

To find the local maximum and minimum points for $f(x) = \frac{3x}{x^2 - 1}$ , follow these steps:

Find the critical points by setting the derivative of $f(x)$ equal to zero and solving for $x$ .
Determine the second derivative of $f(x)$ to classify the critical points as local maxima, minima, or points of inflection.
Evaluate $f(x)$ at the critical points and endpoints of the domain to find the local maximum and minimum values.

Would you like a detailed explanation of each step?

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.