How do you find $(f g) (5) + f (4)$ $(fg)(5)+f(4)$ when $f (x) = (x^{2}) + 1$ $f(x)=(x^2)+1$ and $g (x) = x - 4$ $g(x)=x-4$ ?

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Answer 1 · 2015-12-25T01:03:51

Answer: 43

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

$g (x) = x - 4$ $g(x) = x-4$

To find $(f g) (5) + f (4)$ $(fg)(5) + f(4)$

Step 1: Find $f (4)$ $f(4)$

$f (4) = 4^{2} + 1 = 17$ $f(4) = 4^2 + 1 = color(red)17$

Step 2: $(f g) (x) = (x^{2} + 1) (x - 4)$ $(fg)(x) = (x^2 +1)(x-4)$

$(f g) (5) = (5^{2} + 1) (5 - 4) = (26) (1) = 26$ $(fg)(5) = (5^2 +1)(5-4)= (26)(1) = color(blue)26$

so $(f g) (5) + f (4) = 26 + 17 = 43$ $(fg)(5)+f(4) =color(blue)( 26 )+ color(red)(17) =43$

How do you find (fg)(5)+f(4)(fg)(5)+f(4) when f(x)=(x2)+1f(x)=(x^2)+1 and g(x)=x−4g(x)=x-4?