# Question #a0d1e

Mar 19, 2017

$f \left(5\right) = 7$

#### Explanation:

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

$f \left(2 + 5\right) = 2 {\left(2\right)}^{2} - 3 \left(2\right) + 5 = 8 - 6 + 5 = 7$

Mar 19, 2017

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

$f \left(5\right) = 7$

#### Explanation:

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

$g \left({g}^{-} 1 \left(x\right)\right) = {g}^{-} 1 \left(x\right) + 3$

$x = {g}^{-} 1 \left(x\right) + 3$

${g}^{-} 1 \left(x\right) = x - 3$

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

$f \left(x - 3 + 3\right) = 2 {\left(x - 3\right)}^{2} - 3 \left(x - 3\right) + 5$

$f \left(x\right) = 2 \left({x}^{2} - 6 x + 9\right) - 3 x + 9 + 5$

$f \left(x\right) = 2 {x}^{2} - 12 x + 18 - 3 x + 14$

$f \left(x\right) = 2 {x}^{2} - 15 x + 32$

Evaluate at $x = 5$:

$f \left(5\right) = 2 {\left(5\right)}^{2} - 15 \left(5\right) + 32$

$f \left(5\right) = 7$

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

$f \left(2 + 3\right) = 2 {\left(2\right)}^{2} - 3 \left(2\right) + 5$

$f \left(5\right) = 7$