How do you find $(g o f) (x)$ $(gof)(x)$ given $f (x) = x - 1$ $f(x)=x-1$ and $g (x) = x^{2} + 2 x - 8$ $g(x)=x^2+2x-8$ ?

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Answer 1 · 2017-08-02T23:15:41

Another notation is given by the following equation:

$(g o f) (x) = g (f (x))$ $(gof)(x) = g(f(x))$

We substitute $g (x) = x^{2} + 2 x - 8$ $g(x) = x^2 + 2x -8$ on the right side but we write $(f (x))$ $(f(x))$ everywhere that there is an x:

$(g o f) (x) = {(f (x))}^{2} + 2 (f (x)) - 8$ $(gof)(x) = (f(x))^2 + 2(f(x)) -8$

Substitute $x - 1$ $x-1$ for $f (x)$ $f(x)$ :

$(g o f) (x) = {(x - 1)}^{2} + 2 (x - 1) - 8$ $(gof)(x) = (x-1)^2 + 2(x-1) -8$

Expand the square:

$(g o f) (x) = x^{2} - 2 x + 1 + 2 (x - 1) - 8$ $(gof)(x) = x^2-2x + 1 + 2(x-1) -8$

Use the distributive property:

$(g o f) (x) = x^{2} - 2 x + 1 + 2 x - 2 - 8$ $(gof)(x) = x^2-2x + 1 + 2x-2 -8$

Combine like terms:

$(g o f) (x) = x^{2} - 9 \leftarrow$ $(gof)(x) = x^2 -9 larr$ answer

How do you find (gof)(x)(gof)(x) given f(x)=x−1f(x)=x-1 and g(x)=x2+2x−8g(x)=x^2+2x-8?