Problem 33
Question
Sketch the graph of the function.\(f(x)=e^{2 x}\)
Step-by-Step Solution
Verified Answer
The graph of the function \(f(x) = e^{2x}\) is an upward curve in the first quadrant, passing through the point (0,1), and it approaches, but never reaches, the x-axis as \(x\) decreases.
1Step 1: Understand the Function
The given function \(f(x)=e^{2x}\) is an exponential function. We know that \(e^{2x}\) grows faster than \(e^{x}\) for positive values of \(x\) and decreases to zero much quicker for negative values of \(x\). This should be reflected in our graph.
2Step 2: Identify Key Points
Firstly, for \(x=0\), we have \(f(0)=e^{2*0}=1\), the y-intercept. Then, choose some values of \(x\) where \(x>0\), for instance, \(x=1\), and \(x=2\), and we have \(f(1)=e^{2}=7.39\), and \(f(2)=e^{4}=54.60\) respectively. For \(x<0\), choose \(x=-1\), and \(x=-2\), then we get \(f(-1)= e^{-2}=0.14\) and \(f(-2)=e^{-4}=0.018\), respectively.
3Step 3: Sketch the Graph
Draw the Cartesian coordinate system and mark the key points calculated in the previous step. Then curve upwards in the positive direction along these plotted points, which reflects the nature of our exponential function \(f(x)=e^{2x}\). For \(x<0\), the curve should approach the x-axis but never touch or cross it, this is the asymptote of the function.
Key Concepts
Graph SketchingKey Points IdentificationAsymptotes
Graph Sketching
Graph sketching is all about capturing the essence of a function visually. When dealing with exponential functions like \(f(x) = e^{2x}\), it's important to observe the rapid growth for positive values of \(x\) and swift approach to zero for negative values.
Start by drawing a set of axes. The horizontal axis is your \(x\)-axis, and the vertical one is the \(y\)-axis. Choose appropriate scales based on the values you have.
With exponential functions, you'll quickly notice steep growth.
Start by drawing a set of axes. The horizontal axis is your \(x\)-axis, and the vertical one is the \(y\)-axis. Choose appropriate scales based on the values you have.
With exponential functions, you'll quickly notice steep growth.
- For positive \(x\), the function moves upwards rapidly.
- For negative \(x\), the line approaches closer to but never touches the \(x\)-axis.
Key Points Identification
Identifying key points is crucial when sketching graphs. It allows you to capture pivotal values that define your graph's shape. For the function \(f(x) = e^{2x}\), start with the y-intercept which occurs at \(x = 0\).
At this point, \(f(0) = e^0 = 1\) — a critical anchor on your graph.
At this point, \(f(0) = e^0 = 1\) — a critical anchor on your graph.
- For \(x = 1\), calculate \(f(1) = e^2 \approx 7.39\).
- For \(x = 2\), calculate \(f(2) = e^4 \approx 54.60\).
- For \(x = -1\), calculate \(f(-1) = e^{-2} \approx 0.14\).
- For \(x = -2\), calculate \(f(-2) = e^{-4} \approx 0.018\).
Asymptotes
An asymptote of a function is a line that the graph of the function approaches but never touches.
For an exponential function like \(f(x) = e^{2x}\), the horizontal line \(y = 0\) serves as the asymptote.
This signifies that as \(x\) reduces and approaches negative infinity, \(f(x)\) gets infinitely closer to \(y=0\), but it never actually reaches zero.
Asymptotes are essential in understanding the boundary behavior of graphs:
For an exponential function like \(f(x) = e^{2x}\), the horizontal line \(y = 0\) serves as the asymptote.
This signifies that as \(x\) reduces and approaches negative infinity, \(f(x)\) gets infinitely closer to \(y=0\), but it never actually reaches zero.
Asymptotes are essential in understanding the boundary behavior of graphs:
- They guide the sketching of the graph, ensuring it never crosses certain regions.
- They help visualize limits. In this case, \( e^{2x} \) approaches zero but never becomes negative.
Other exercises in this chapter
Problem 33
Evaluate the logarithm. Round your result to three decimal places.\(\log _{1 / 4} 10\)
View solution Problem 33
Evaluate the expression without using a calculator.\(\log _{7} 7\)
View solution Problem 34
Solve the exponential equation algebraically. Approximate the result to three decimal places.\(2^{x-3}=31\)
View solution Problem 34
Evaluate the logarithm. Round your result to three decimal places.\(\log _{1 / 3} 5\)
View solution