If $f (x) = \frac{x}{4} - 3$ $f(x)=x/4-3$ and $g (x) = 4 x^{2} + 2 x - 4$ $g(x)=4x^2+2x-4$ find $(f + g) (x)$ $(f+g)(x)$ ?

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Answer 1 · 2018-06-15T16:55:29

$(f + g) (x) = 4 x^{2} + \frac{9}{4} x - 7$ $(f+g)(x) = 4x^2+9/4x-7$

$(f + g) (x) = f (x) + g (x)$ $(f+g)(x) = f(x)+g(x)$

Substitute $f (x) = \frac{x}{4} - 3$ $f(x)=x/4-3$ :

$(f + g) (x) = \frac{x}{4} - 3 + g (x)$ $(f+g)(x) = x/4-3+g(x)$

Substutitute $g (x) = 4 x^{2} + 2 x - 4$ $g(x)=4x^2+2x-4$ :

$(f + g) (x) = \frac{x}{4} - 3 + 4 x^{2} + 2 x - 4$ $(f+g)(x) = x/4-3+4x^2+2x-4$

Combine like terms:

$(f + g) (x) = 4 x^{2} + \frac{9}{4} x - 7$ $(f+g)(x) = 4x^2+9/4x-7$

If f(x)=x/4-3f(x)=x4−3f(x)=x/4-3 and g(x)=4x^2+2x-4g(x)=4x2+2x−4g(x)=4x^2+2x-4 find (f+g)(x)(f+g)(x)(f+g)(x) ?