What are the domain and range of the function y=2(3^x)-1?

  • Thread starter Thread starter aisha
  • Start date Start date
  • Tags Tags
    Domain Range
AI Thread Summary
The function y=2(3^x)-1 has a domain of all real numbers, meaning any x-value can be used. The range is the open interval (-1, +∞), indicating that the function outputs values greater than -1. The x-intercept is approximately -0.63, and the y-intercept is 1, confirming where the function intersects the axes. The vertical asymptote is at -0.63, while the horizontal asymptote is at y=0, showing the behavior of the function as x approaches these limits. Understanding these concepts is crucial for analyzing the function's characteristics effectively.
aisha
Messages
584
Reaction score
0
I was given the equation y=2(3^x)-1 and told to state the domain, range, x and y intercepts and asymptotes.

I graphed the eqn and got the x-intercept to =-0.63 and the y-intercept=1
the vertical asymptote is the same as x-intercept so =-0.63 and the horizontal asymptote always equals 0.

If these values are right, then I need to find out the domain and range, but I don't know how. Can someone please help me? :redface:
 
Physics news on Phys.org
aisha said:
I was given the equation y=2(3^x)-1 and told to state the domain, range, x and y intercepts and asymptotes.
I graphed the eqn and got the x-intercept to =-0.63 and the y-intercept=1
the vertical asymptote is the same as x-intercept so =-0.63 and the horizontal asymptote always equals 0.
If these values are right, then I need to find out the domain and range, but I don't know how. Can someone please help me? :redface:


Me again,...This time i say to review calculations.
Hints:1)Do u agree with me that the domain in the entire real axis and the range the open interval (-1,+infinity)??
2)Use the correct definitions of veritical/horizontal asymptotes.
3) Solve the equations for the intersections correctly.

PS I assumed that your initial function was y=2\cdot 3^{x} -1.
 


The domain of a function is the set of all possible input values, or x-values, for which the function is defined. In this case, the function y=2(3^x)-1 is defined for all real numbers, so the domain is all real numbers, or (-∞, ∞). This means that any x-value can be plugged into the function and it will produce a valid output.

The range of a function is the set of all possible output values, or y-values, that the function can produce. In this case, the function y=2(3^x)-1 can produce any real number as an output, so the range is also all real numbers, or (-∞, ∞). This means that the function can have a y-value of any number, positive or negative, depending on the input x-value.

As for the x and y-intercepts, your calculations are correct. The x-intercept is approximately -0.63 and the y-intercept is 1. These values represent the points where the function crosses the x and y-axes, respectively.

The vertical asymptote, as you mentioned, is also located at -0.63. This is the value where the function approaches infinity as x approaches -0.63 from either side. The horizontal asymptote, as you also correctly stated, is always at y=0. This means that as x increases or decreases without bound, the function will approach 0 as well.

I hope this helps clarify the concepts of domain, range, x and y-intercepts, and asymptotes for this function. Keep up the good work!
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Thread 'A bead-mass oscillatory system problem'
I can't figure out how to find the velocity of the particle at 37 degrees. Basically the bead moves with velocity towards right let's call it v1. The particle moves with some velocity v2. In frame of the bead, the particle is performing circular motion. So v of particle wrt bead would be perpendicular to the string. But how would I find the velocity of particle in ground frame? I tried using vectors to figure it out and the angle is coming out to be extremely long. One equation is by work...
Back
Top