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How do you find #(f/g)(2)# given #f(x)=x^2-1 and #g(x)=2x-3# and #h(x)=1-4x#?

Answer 1

Ethan Adkins

((f/g)(2) = \frac{f(2)}{g(2)} = \frac{(2^2 - 1)}{(2 \cdot 2 - 3)} = \frac{3}{1})

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Answer 2

Avery Alexander

See a solution process below:

First, we can write #(f/g)(x)# as:

#(f/g)(x) = (x^2 - 1)/(2x - 3)#

We can now substitute #color(red)(2)# for each occurrence of color(red)(x)# in #(f/g)(x)# and calculate the result:

#(f/g)(color(red)(x)) = (color(red)(x)^2 - 1)/(2color(red)(x) - 3)# becomes:

#(f/g)(color(red)(2)) = (color(red)(2)^2 - 1)/((2 * color(red)(2)) - 3)#

#(f/g)(color(red)(2)) = (4 - 1)/(4 - 3)#

#(f/g)(color(red)(2)) = 3/1#

#(f/g)(color(red)(2)) = 3#

The #h(x)# function is extraneous information in this problem.

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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.