Q. 112 - Precalculus Enhanced with Graphing Utilities

Question

^{Suppose that $F\left(x\right)={\log }_2\left(x+1\right)-3.$}

(a) What is the domain of $F?$

(b) What is $F\left(7\right)?$ What point is on the graph of $F?$ 

(c) If $F\left(x\right)=-1,$ what is $x?$ What point is on the graph of $F?$

(d) What is the zero of $F?$

Step-by-Step Solution

Answer

(a) The domain of $F$ is $\left(-1,\infty \right).$

(b) The value of $F\left(7\right)=0$ and the point in the graph is $\left(7,0\right).$

(c) The value of $x$ is $3$ and the point in the graph is $\left(3,-1\right).$

(d) The zero of $F$ is $7.$

Part (a) Step 1. Given information

It is given that $F\left(x\right)={\log }_2\left(x+1\right)-3.$ We need to determine the domain of $F.$

Step 2. Determine the domain

We know, in function $f\left(a\right)={\log }_b^a,a>0.$

Here,

         $a=x+1.$

     $x+1>0.$

$x+1-1>0-1.$

            $x>-1.$

Thus, the domain of the given function is $\left(-1,\infty \right).$

Part (b) Step 1. Given information

It is given that $F\left(x\right)={\log }_2\left(x+1\right)-3.$ We need to determine $F\left(7\right)$ and  point the  graph of $F.$

Step 2. Simplify F 7

Substitute $7$ to $x$ in $F\left(x\right)={\log }_2\left(x+1\right)-3.$

$F\left(7\right)={\log }_2\left(7+1\right)-3.$

       $={\log }_{2}8-3.$

Since $8=2^3,$ then ${\log }_{2}8=3.$

Thus $F\left(7\right)={\log }_{2}8-3.$

                 $=3-3.$

                 $=0.$

Step 3. Determining the point on graph F x

We know, if $f\left(a\right)=b,$ then the point $\left(a,b\right)$ is on the graph of $f\left(x\right).$ Since $F\left(7\right)=0,$ then the point $\left(7,0\right)$ is on the graph of $F.$

Part (c) Step 1. Given information

It is given that $F\left(x\right)={\log }_2\left(x+1\right)-3.$ We need to determine the value of $x$ in $f\left(x\right)=-1,$ and the point is the graph.

Step 2. Determine the point

Substitute $-1$ to $F\left(x\right)={\log }_2\left(x+1\right)-3.$

     $-1={\log }_2\left(x+1\right)-3.$

$-1+3={\log }_2\left(x+1\right).$

         $2={\log }_2\left(x+1\right).$

Use the rule $a={\log }_{b}c\Rightarrow b^2=c.$

    $2^2=x+1.$

      $4=x+1.$

$4-1=x+1-1.$

      $3=x.$

The point of the graph is $\left(3,-1\right)$ is on the graph of $F.$

Part (d) Step 1. Given information

It is given that $F\left(x\right)={\log }_2\left(x+1\right)-3.$ We need to determine zero of $x.$

Step 2. Simplify

$F\left(x\right)=0.$

     $0={\log }_2\left(x+1\right)-3.$

     $3={\log }_2\left(x+1\right).$

    $2^3=x+1.$

     $8=x+1.$

     $x=7.$

The zero of $F$ is $7.$