Determine the growth or decay of a function with a rational exponent Precalculus

We have g\left(x\right)= \dfrac{1}{\sqrt{2x+3}}

The composition of an increasing function with another increasing function is increasing on its domain. We have:

• f_1\left(x\right)= 2x+3
• f_2\left(x\right)= \sqrt{x}

The two functions are increasing. Hence the following function is also increasing:

h\left(x\right)=\left(f_2 \circ f_1\right)\left(x\right) = \sqrt{2x+3}

On the other hand, the composition of an increasing function with a decreasing function is decreasing on its domain. Consider the decreasing function:

f_3\left(x\right) = \dfrac{1}{x}

We have:

\left(f_3 \circ h \right)\left(x\right) = \dfrac{1}{\sqrt{2x+3}}

It is decreasing on its domain.

We have g\left(x\right)= \sqrt[3]{5x-9}

The composition of an increasing function with another increasing function is increasing on its domain. We have:

• f_1\left(x\right)=\sqrt[3]{x}
• f_2\left(x\right)= 5x-9

The two functions are increasing. Hence the following function is also increasing:

g\left(x\right)=\left(f_1 \circ f_2\right)\left(x\right) = \sqrt[3]{5x-9}

It is increasing on its domain.

We have g\left(x\right)= \dfrac{1}{\sqrt[4]{\left(2x+1\right)^{3}}}

The composition of an increasing function with another increasing function is increasing on its domain. We have:

• f_1\left(x\right)= 2x+1
• f_2\left(x\right)= x^{3}
• f_3\left(x\right)= \sqrt[4]{x}

The three functions are increasing. Hence the following function is also increasing:

h\left(x\right)=\left(f_3 \circ f_2 \circ f_1\right)\left(x\right) = \sqrt[4]{\left(2x+1\right)^{3}}

On the other hand, the composition of an increasing function with a decreasing function is decreasing on its domain. Consider the decreasing function:

f_4\left(x\right) = \dfrac{1}{x}

We have:

\left(f_4 \circ h \right)\left(x\right) = \dfrac{1}{\sqrt[4]{\left(2x+1\right)^{3}}}

It is decreasing on its domain.

We have g\left(x\right)= \sqrt{x-3}

The composition of an increasing function with another increasing function is increasing on its domain. We have:

• f_1\left(x\right)=\sqrt{x}
• f_2\left(x\right)= x-3

The two functions are increasing. Hence the following function is also increasing:

g\left(x\right)=\left(f_1 \circ f_2\right)\left(x\right) = \sqrt{x-3}

It is increasing on its domain.

We have g\left(x\right)= \dfrac{1}{\sqrt[4]{\ln\left(x\right)-3}}

The composition of an increasing function with another increasing function is increasing on its domain. We have:

• f_1\left(x\right)= \ln\left(x\right)
• f_2\left(x\right)= x-3
• f_3\left(x\right)= \sqrt[4]{x}

The three functions are increasing. Hence the following function is also increasing:

h\left(x\right)=\left(f_3 \circ f_2 \circ f_1\right)\left(x\right) = \sqrt[4]{\ln\left(x\right)-3}

On the other hand, the composition of an increasing function with a decreasing function is decreasing on its domain. Consider the decreasing function:

f_4\left(x\right) = \dfrac{1}{x}

We have:

\left(f_4 \circ h \right)\left(x\right) = \dfrac{1}{\sqrt[4]{\ln\left(x\right)-3}}

It is decreasing on its domain.

We have g\left(x\right)= \sqrt[3]{3e^{x}+5}

The composition of an increasing function with another increasing function is increasing on its domain. We have:

• f_1\left(x\right)= e^{x}
• f_2\left(x\right)= 3x+5
• f_3\left(x\right)= \sqrt[3]{x}

The three functions are increasing. Hence the following function is also increasing:

g\left(x\right)=\left(f_3 \circ f_2 \circ f_1\right)\left(x\right) = \sqrt[3]{3e^{x}+5}

It is increasing on its domain.

We have g\left(x\right)= \dfrac{1}{\sqrt[5]{\left(3x-4\right)^{3}}}

The composition of an increasing function with another increasing function is increasing on its domain. We have:

• f_1\left(x\right)= 3x-4
• f_2\left(x\right)= x^{3}
• f_3\left(x\right)= \sqrt[5]{x}

The three functions are increasing. Hence the following function is also increasing:

h\left(x\right)=\left(f_3 \circ f_2 \circ f_1\right)\left(x\right) = \sqrt[5]{\left(3x-4\right)^{3}}

On the other hand, the composition of an increasing function with a decreasing function is decreasing on its domain. Consider the decreasing function:

f_4\left(x\right) = \dfrac{1}{x}

We have:

\left(f_4 \circ h \right)\left(x\right) = \dfrac{1}{\sqrt[5]{\left(3x-4\right)^{3}}}

It is decreasing on its domain.