# If f(n) = 9^-n, then what is f(-1/4)?

Feb 20, 2016

$= \sqrt{3}$

#### Explanation:

put $- \frac{1}{4}$ in place of n.

$f \left(- \frac{1}{4}\right)$

$= {9}^{- \left(- \frac{1}{4}\right)}$

$= {9}^{\frac{1}{4}}$

$= {3}^{2 \cdot \left(\frac{1}{4}\right)}$

$= {3}^{\frac{1}{2}}$

$= \sqrt{3}$

Feb 20, 2016

The number in the parentheses to the right replaces $n$. That's to say you must plug the number in the parentheses in place of n in the equation

#### Explanation:

$f \left(n\right) = {9}^{-} n \to n = - \frac{1}{4}$

f(-1/4) = 9^(-(-1/4)

$f \left(- \frac{1}{4}\right) = {9}^{\frac{1}{4}}$

$f \left(- \frac{1}{4}\right) = \sqrt{9}$

$f \left(- \frac{1}{4}\right) = \sqrt{3}$

When they give you the equation and something like $f \left(x\right) = 4$, it means they give you the value of y and want you to find the value of x. When they replace x in the parentheses with a number and they give the equation, you're looking for y.

Practice exercises:

1. Evaluate, assuming the equations are of the form $f \left(x\right) = y$

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

b) $f \left(- 3\right) = \sqrt{- 4 x + 4}$

c) $f \left(x\right) = 3 x - 11 , f \left(x\right) = 16$

d). $f \left(x\right) = {3}^{x} , f \left(x\right) = \frac{1}{27}$

Good luck!