# Is the function f(x) = (18)^x increasing or decreasing?

Oct 28, 2015

depend on the value of $x$. See explanation

#### Explanation:

if $x \to x > 1$ then ${18}^{x}$ is increasing as $x$ increases

if $x \to 0 < x < 1$ then ${18}^{x}$ is decreasing as $x$ approaches 0 from the positive side.

If $x$ is negative then we have $\frac{1}{{18}^{x}}$ so:

If $x \to \left(- 1\right) < x < 0$ then as $x$ approaches 0 the ${18}^{x}$ in $\frac{1}{{18}^{x}}$ gets smaller and smaller so $\frac{1}{18} ^ x$ gets bigger and bigger

If $x \to x < \left(- 1\right)$ then the ${18}^{x}$ in $\frac{1}{{18}^{x}}$ gets bigger and bigger as $x$ becomes increasingly les -1. So $\frac{1}{{18}^{x}}$ becomes less and less.

Oct 30, 2015

The function is increasing.

#### Explanation:

An exponential function $y = {a}^{x}$ is :

• increasing for $a > 1$
• decreasing for a in (0;1)
• not defined for $a < 0$ (because negative numbers cannot be raised to rational powers ${\left(- 1\right)}^{\frac{1}{2}} = \sqrt{- 1}$ is not a real number)