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Answer 1 · 2017-03-19T16:02:00

$f(5) = 7$

$f(5) = f(2+3)$ which means $x = 2$

$f(2+5) = 2(2)^2 - 3(2) + 5 = 8 - 6 + 5 = 7$

Answer 2 · 2017-03-19T16:16:38

There are at least two ways to do this problem; please see the explanation for both.

$f(5)=7$

Let $g(x) = x + 3$ , then:

$g(g^-1(x))= g^-1(x)+3$

$x = g^-1(x)+3$

$g^-1(x)=x-3$

Therefore, we can obtain $f(x)$ by substituting $x-3$ for $x$ into $f(x+3)$

$f(x-3+3)=2(x-3)^2-3(x-3)+5$

$f(x) = 2(x^2-6x+9) - 3x+9+5$

$f(x) = 2x^2-12x+18-3x+14$

$f(x) = 2x^2-15x+32$

Evaluate at $x = 5$ :

$f(5) = 2(5)^2-15(5)+32$

$f(5) = 7$

Alternatively we could have evaluated $f(x+3)$ at $x = 2$ :

$f(2+3)= 2(2)^2 -3(2)+5$

$f(5)= 7$