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

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

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Answer 1 · 2017-04-10T12:01:55

See the entire solution process below:

Explanation:

First,

$(f + g) (x) = f (x) + g (x) = x^{2} - 1 + 2 x - 3$ $(f + g)(x) = f(x) + g(x) = x^2 - 1 + 2x - 3$

We can ignore $h (x)$ $h(x)$ as it is extraneous information to this problem.

To find $(f + g) (- 2)$ $(f + g)(-2)$ we need to substitute $- 2$ $color(red)(-2)$ for each occurrence of $x$ $color(red)(x)$ in $(f + g) (x)$ $(f + g)(x)$ :

$(f + g) (x) = x^{2} - 1 + 2 x - 3$ $(f + g)(color(red)(x)) = color(red)(x)^2 - 1 + 2color(red)(x) - 3$ becomes:

$(f + g) (- 2) = {(- 2)}^{2} - 1 + (2 \cdot - 2) - 3$ $(f + g)(color(red)(-2)) = (color(red)(-2))^2 - 1 + (2 * color(red)(-2)) - 3$

$(f + g) (- 2) = 4 - 1 - 4 - 3$ $(f + g)(color(red)(-2)) = 4 - 1 - 4 - 3$

$(f + g) (- 2) = - 4$ $(f + g)(color(red)(-2)) = -4$

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

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Explanation:

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How do you find (f+g)(−2)(f+g)(-2) given f(x)=x2−1andf(x)=x^2-1 and g(x)=2x-3and and h(x)=1-4x#?

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Explanation:

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