Question

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

(a) What is $F(-5)$? What point is on the graph ofF ?

(b) If $F\left(x\right)=24$, what is x? What point is on the graph of F ?

(c) Find the zero of F.

Short answer

Part (a) $F(-5)=240$ and the point on the graph of F is $(-5,240)$.

Part (b) $x=-3$ and the point on the graph of F is $(-3,24)$.

Part (c) The zero of F is$x=-1$.

Step-by-step solution

Part (a) Step 1. Substitute x=-5 in F(x)=(13)x-3.

This gives

$$\begin{gathered} F(-5)={\left(\frac{1}{3}\right)}^{-5}-3 \\ =3^5-3 \\ =243-3 \\ =240 \end{gathered}$$

The point on the graph of the function F is$(-5,240)$.

Part (b) Step 2. Finding x using F(x)=24. 

We have

$$\begin{gathered} F\left(x\right)=24 \\ {\left(\frac{1}{3}\right)}^x-3=24 \\ {\left(\frac{1}{3}\right)}^x=24+3 \\ {3}^{-x}=27 \\ {3}^{-x}=3^3 \end{gathered}$$

On comparing,

$x=-3$

The point on the graph of the function F is$(-3,24)$.

Part (c) Step 1. Substitute F(x)=0

This gives

$$\begin{gathered} {\left(\frac{1}{3}\right)}^x-3=0 \\ {\left(\frac{1}{3}\right)}^x=3 \\ {3}^{-x}=3^1 \end{gathered}$$

On comparing,

$x=-1$