Question

Let f(x) = -2x^2 -x, and let g(x) = \frac{f(x)-f(-1)}{x+1}, x \ne -1. Find g(10).

Answer 1

The value of g(10) is -19

How to determine the value

Given the functions:

• f(x) = -2x^2 -x
• g(x) = \frac{f(x)-f(-1)}{x+1}

To find g(10), let's substitute the expressions into the function;

g(x) = $\frac{f(x) - f(-1) }{x + 1}$

Let's find f(-1)

f(x) = - 2x^2 -x

f(-1) = -2 (-1)² - (-1)

f(-1) = -2(1) + 1

f(-1) = -1

substitute the value of f(-1) as - 1

g(x) = $\frac{-2x^2 - x }{x + 1}$

g(10) = $\frac{-2(10)^2 - (10) - (-1)}{10 + 1}$

g(10) = $\frac{-2(100)- 10 + 1}{11}$

g(10) = $\frac{-200 - 9}{11}$

g(10) = $\frac{-209}{11}$

g(10) = - 19

Thus, the value of g(10) is -19

Learn more about functions here:

brainly.com/question/2328150

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