Respuesta :
Answer:
Option (C) is correct.
Step-by-step explanation:
The given options of the possible equation for the graph are as follows:
[tex](A) y=2\left(\frac{3}{2}\right)^x \\\\(B) y=-2\left(\frac{3}{2}\right)^{-x} \\\\(C) y=2\left(\frac{2}{3}\right)^x \\\\(D) y=-2\left(\frac{2}{3}\right)^{-x} \\\\[/tex]
The given graph is decreasing and at x=0, y=2.
So, first checking the value of the given options for x=0
[tex](A) y=2\left(\frac{3}{2}\right)^0=2\times 1= 2 \\\\(B) y=--2\left(\frac{3}{2}\right)^{-0}= -2\times 1= -2 \; (not\; possible) \\\\(C) y=2\left(\frac{2}{3}\right)^0= 2\times 1= 2 \\\\(D) y=2\left(\frac{2}{3}\right)^{-0} = -2\times 1= -2 \; (not\; possible)[/tex]
As, for x=0, y=2, so options (C) and (D) are not possible, so rejected.
Now, checking the nature (increasing or decreasing) of the given equation by differentiating it.
For option (A),
[tex]\frac{dy}{dx}=2\left(\frac{3}{2}\right)^{x}\times \ln\left(\frac{3}{2}\right)[/tex]
As [tex]\ln \left(\frac{3}{2}\right)=\ln(1.5)>0 \;and\; \left(\frac{3}{2}\right)^{x} >0[/tex]
So, [tex]\frac{dy}{dx}>0[/tex]
Therefore, the function in option (A) is increasing function.
Similarly, for option (C),
[tex]\frac{dy}{dx}=2\left(\frac{2}{3}\right)^{x}\times \ln\left(\frac{2}{3}\right)[/tex]
As [tex]\ln \left(\frac{2}{3}\right)=\ln(0.67)<0 \;and\; \left(\frac{3}{2}\right)^{x} >0[/tex]
So, [tex]\frac{dy}{dx}<0[/tex]
Therefore, the function in option (C) is decreasing function.
As the given graph is decreasing, so, (C) represents[tex]y=2\left(\frac{2}{3}\right)^x[/tex] the given graph.
Hence, option (C) is correct.
The equation [tex](C) y = 2(\dfrac{2}{3})^{x[/tex] could be the equation to represent the given graph is decreasing.
We have to determine, Which of the following equations could be the equation to represent the given graph.
According to the question,
The given options of the possible equation for the graph are as follows:
- [tex](A)y = 2(\frac{3}{2})^x[/tex]
- [tex](B)y = -2(\frac{3}{2})^{-x}[/tex]
- [tex](C)y = 2(\frac{2}{3})^{x}[/tex]
- [tex](D)y = -2(\frac{2}{3})^{-x}[/tex]
The given graph is decreasing and at x = 0, y = 2.
Then,
First checking the value of the given options for x = 0.
- [tex](A)y = 2(\frac{3}{2})^x = 2(\frac{3}{2})^0 = 2 \times1 = 2[/tex]
- [tex](B)y = -2(\frac{3}{2})^{-x} = (A)y = -2(\frac{3}{2})^{-0} = -2 \times 1 = -2 (not \ possible)[/tex]
- [tex](C)y = 2(\frac{2}{3})^x = 2(\frac{2}{3})^0 = 2 \times1 = 2[/tex]
- [tex](D)y = -2(\frac{2}{3})^{-x} = -2(\frac{2}{3})^{-0} = -2 \times1 = -2 (not \ possible)[/tex]
For x = 0, y = 2, so options (C) and (D) are not possible, so the options are rejected.
Therefore,
To check the nature (increasing or decreasing) of the given equation by differentiating it.
- For function A;
[tex]\dfrac{dy}{dx} = 2(\dfrac{2}{3})^x \times ln(\dfrac{3}{2})\\\\ln(\dfrac{3}{2}) = ln(1.5) >0 \ and\ (\dfrac{3}{2})^{x} >0\\\\So, \dfrac{dy}{dx}>0\\\\[/tex]
Therefore, The function in option A is increasing function.
For function c;
[tex]\dfrac{dy}{dx} = 2(\dfrac{2}{3})^x \times ln(\dfrac{2}{3})\\\\ln(\dfrac{3}{2}) = ln(0.67) <0 \ and\ (\dfrac{3}{2})^{x} >0\\\\So, \dfrac{dy}{dx}<0\\\\[/tex]
Therefore, The function in option c is decreasing function.
The given graph is decreasing, so, the function c represents the given graph decreasing.
Hence, option (C) is correct.
To know more about the Graphs click the link given below.
https://brainly.com/question/16351335
