Question

Find the following. $f(x) = \sqrt{x^2 - 9}$, $g(x) = \frac{x^2}{x^2 + 1}$ (a) $(f + g)(x) = $ (b) $(f - g)(x) = $ (c) $(fg)(x) = $ (d) $(f/g)(x) = $ What is the domain of $f/g$? (Enter your answer using interval notation.)

          Find the following.
$f(x) = \sqrt{x^2 - 9}$, $g(x) = \frac{x^2}{x^2 + 1}$
(a) $(f + g)(x) = $
(b) $(f - g)(x) = $
(c) $(fg)(x) = $
(d) $(f/g)(x) = $
What is the domain of $f/g$? (Enter your answer using interval notation.)

Find the following.
f(x) = √(x^2 - 9), g(x) = (x^2)/(x^2 + 1)
(a) (f + g)(x) =
(b) (f - g)(x) =
(c) (fg)(x) =
(d) (f/g)(x) =
What is the domain of f/g? (Enter your answer using interval notation.)

Added by Victoria W.

Precalculus with Limits

Ron Larson 2nd Edition

Instant Answer

Solved by Expert Anurag Kumar

Step 1

(a) (f + g)(x) = f(x) + g(x) = √x - 9 + x = x + √x - 9 (b) (f - g)(x) = f(x) - g(x) = √x - 9 - x = -x + √x - 9 (c) (fg)(x) = f(x)g(x) = (√x - 9)x = x√x - 9x (d) (f/g)(x) = f(x)/g(x) = (√x - 9)/(x), but we need to consider the domain of g(x) which is all real Show more…

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Find the following: f(x) = âˆšx - 9, g(x) (a) (f + g)(x) (b) (f - g)(x) (c) (f * g)(x) (d) (f âˆ˜ g)(x) What is the domain of f âˆ˜ g? (Enter your answer using interval notation)