Let f(x)=e^(2x)-kx, for k greater than 0.
Using a calculator or computer, sketch the graph of f for k=1/9, 1/6,1/3,1/2,1,2,4. Describe what happens as k changes.
f(x) has a local minimum. Find the location of the minimum.
x= ____
Find the y-coordinate of the minimum.
y= _____
Find the value of k for which this y-coordinate is largest.
k= ______
How do you know that this value of k maximizes the y-coordinate? Find d^2y/dk^2 to use the second-derivative test.
d^2y/d^2k=
(Note that the derivative you get is negative for all positive values of k, and confirm that you agree that this means that your value of k maximizes the y-coordinate of the minimum.)
MY ANSWERS ARE=
Loc. min= 1
y-coordinate=1
k where y is largest= 2
d^2y/d^2k = 4e^(2x)
The only one I got right plugging in these values was k when y is largest...
I'm sorry if my previous response didn't match with your answers. Let's go through the questions again:
1. Sketching the graph of f(x) for different values of k:
- k = 1/9: The graph will have a steep curve and start from a high positive value at x = 0, then gradually decrease.
- k = 1/6: The graph will have a slightly less steep curve compared to the previous one, still starting from a high positive value at x = 0.
- k = 1/3: The graph will have a less steep curve than before, starting from a positive value at x = 0.
- k = 1/2: The graph will have a moderate slope and start from a positive value at x = 0.
- k = 1: The graph will have a gentler curve compared to the previous ones, still starting from a positive value at x = 0.
- k = 2: The graph will have a steeper curve than k = 1, starting from a higher positive value at x = 0.
- k = 4: The graph will have a very steep curve and start from an even higher positive value at x = 0.
2. Finding the location of the local minimum:
To find the location of the minimum, we need to find the x-coordinate where the derivative of f(x) is equal to 0.
Taking the derivative of f(x), we get f'(x) = 2e^(2x) - k. Setting this equal to 0, we have 2e^(2x) - k = 0.
Solving for x, we find x = ln(k/2) / 2.
3. Finding the y-coordinate of the minimum:
Plugging the x-coordinate found in the previous step back into the original function f(x), we find the y-coordinate.
Substituting x = ln(k/2) / 2 into f(x) = e^(2x) - kx, we get y = e^(ln(k/2)) - k(ln(k/2) / 2) = (k/2) - k(ln(k/2) / 2).
Simplifying further, we have y = k/2 - kln(k/2) / 2.
4. Finding the value of k for which this y-coordinate is largest:
To find the value of k that maximizes the y-coordinate, we need to find the maximum of the function y = k/2 - kln(k/2) / 2.
Taking the derivative of y with respect to k and setting it equal to 0, we have (1/2) - ln(k/2) / 2 = 0.
Solving for k, we find k = 2 * e.
5. Finding d^2y/d^2k to use the second-derivative test:
Taking the second derivative of y with respect to k, we have d^2y/d^2k = -1 / (2k).
This derivative is negative for all positive values of k, which means that the value of k we found in the previous step maximizes the y-coordinate.
So the correct answers are:
- Location of the minimum: x = ln(k/2) / 2
- Y-coordinate of the minimum: y = k/2 - kln(k/2) / 2
- Value of k for which this y-coordinate is largest: k = 2 * e
- d^2y/d^2k = -1 / (2k)
To sketch the graph of f(x) for different values of k, you can use a graphing calculator or computer software to plot the function. Here is a description of what happens as k changes:
1. For k = 1/9: The graph of f(x) will start at a high point and decrease rapidly as x increases. It will approach but never touch the x-axis.
2. For k = 1/6: The graph of f(x) will have a similar shape as k = 1/9, but the function will decrease at a slower rate. It will still approach but never touch the x-axis.
3. For k = 1/3: The graph of f(x) will have a gentler slope compared to the previous cases. It will decrease gradually and eventually touch the x-axis at a specific point.
4. For k = 1/2: The graph of f(x) will have an even gentler slope. It will decrease more slowly and touch the x-axis at a different point compared to k = 1/3.
5. For k = 1: The graph of f(x) will have a horizontal tangent at the point where f(x) touches the x-axis. It will resemble a cubic function with a double root.
6. For k = 2: The graph of f(x) will have a positive slope, but it will still touch the x-axis at a specific point. The shape and orientation of the graph will be similar to the previous cases.
7. For k = 4: The graph of f(x) will start at a low point and increase rapidly as x increases. It will approach but never touch the x-axis.
To find the location of the local minimum, you can use calculus. Take the derivative of f(x) with respect to x:
f'(x) = 2e^(2x) - k
Set f'(x) equal to zero and solve for x:
2e^(2x) - k = 0
2e^(2x) = k
e^(2x) = k/2
2x = ln(k/2)
x = ln(k/2)/2
For the y-coordinate of the minimum, substitute x = ln(k/2)/2 back into the original function:
y = e^(2(ln(k/2)/2)) - k(ln(k/2)/2)
Now, find the value of k for which this y-coordinate is largest:
To maximize the y-coordinate, we need to find the maximum value of the function y with respect to k. Take the derivative of y with respect to k:
y' = -ln(k/2)/4 - ln(k/2)/2 + 1/2
Set y' equal to zero and solve for k:
-ln(k/2)/4 - ln(k/2)/2 + 1/2 = 0
-3ln(k/2)/4 = -1/2
ln(k/2)/4 = 1/2
ln(k/2) = 2
k/2 = e^2
k = 2e^2
To use the second-derivative test, we need to find the second derivative of y with respect to k, which is given by:
d^2y/d^2k = 4e^(2(ln(k/2)/2))
Note that this derivative is always positive for all positive values of k. Therefore, our value of k = 2e^2 does indeed maximize the y-coordinate of the minimum.
So, the revised answers are:
Location of the minimum: x = ln(k/2)/2
Y-coordinate of the minimum: y = e^(2(ln(k/2)/2)) - k(ln(k/2)/2)
Value of k where y is largest: k = 2e^2
Second derivative: d^2y/d^2k = 4e^(2(ln(k/2)/2))
To sketch the graph of f(x)=e^(2x)-kx for different values of k, you can use a calculator or computer to plot the points and connect them to create the graph. Here's how you can do it:
1. Define the function f(x) = e^(2x) - kx.
2. Choose a range of x-values that would be sufficient to capture the behavior of the function. For example, you can choose -2 to 2.
3. For each value of k given (1/9, 1/6, 1/3, 1/2, 1, 2, 4), substitute it into the function and calculate the corresponding y-values for different x-values.
4. Plot the points (x, y) for each value of k on a graph.
5. Connect the points to create the graph.
As for describing what happens as k changes:
- When k is small, such as 1/9, 1/6, and 1/3, the graph of f(x) tends to have a steep slope and approaches positive infinity as x increases.
- As k increases, the graph becomes less steep and attains a point of minimum at some x-value. This point represents the local minimum of the function.
- As k continues to increase, the graph becomes flatter, and the value of the local minimum also increases.
Now let's find the location of the minimum (x-coordinate) and the corresponding y-coordinate for the local minimum.
To find the location of the minimum (x-coordinate):
1. Since f(x) has a local minimum, we need to find where the derivative of f(x) equals zero.
2. Calculate the derivative of f(x) with respect to x: f'(x) = 2e^(2x) - k.
3. Set f'(x) = 0 and solve the equation for x.
2e^(2x) - k = 0
2e^(2x) = k
e^(2x) = k/2
2x = ln(k/2)
x = ln(k/2) / 2
The location of the minimum is x = ln(k/2) / 2.
To find the y-coordinate of the minimum:
1. Substitute the x-coordinate we found into the original function f(x).
f(x) = e^(2 * (ln(k/2) / 2)) - k * (ln(k/2) / 2)
Simplifying further, we get: f(x) = (k/2)^(2/2) - k * (ln(k/2) / 2)
f(x) = (k/2) - k * (ln(k/2) / 2)
The y-coordinate of the minimum is f(x) = (k/2) - k * (ln(k/2) / 2).
To find the value of k for which this y-coordinate is largest:
1. Set up the function for y-coordinate: y = (k/2) - k * (ln(k/2) / 2).
2. Take the derivative of y with respect to k: dy/dk = 1/2 - (ln(k/2) / 2) - (k/2) * (1/(k/2)) * (1/2).
3. Simplify the derivative: dy/dk = 1/2 - (ln(k/2) / 2) - 1/2.
4. Set dy/dk = 0 and solve for k.
0 = 1/2 - (ln(k/2) / 2) - 1/2
ln(k/2) = 0
k/2 = 1
k = 2
The value of k for which the y-coordinate is largest is k = 2.
To find d^2y/dk^2 (the second derivative of y with respect to k):
1. Take the derivative of dy/dk: (d^2y/dk^2) = -1/(k/2).
2. Simplify further: (d^2y/dk^2) = -2/k.
The second derivative is -2/k.
Since the result of the second derivative is negative for all positive values of k, this confirms that the value of k = 2 maximizes the y-coordinate of the minimum. The second derivative test shows that the function has a local maximum at k = 2.