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.)
y = e^(2x)-kx
y' = 2e^(2x) - k
y" = 4e^(2x)
so, now we can answer the questions
y'=0 at x = 1/2 ln(k/2)
y" is always positive, so
there is a minimum at (1/2 ln(k/2) , k/2 (1-ln(k/2))
Now, which value of k makes y=k/2 (1-ln(k/2) a maximum is k=2
since
y' = -1/2 ln(k/2)
y" = -1/2x < 0 at x=2
To sketch the graph of f(x)=e^(2x)-kx for different values of k, we can use a calculator or computer software that can plot graphs.
First, let's evaluate f(x) for k=1/9, 1/6, 1/3, 1/2, 1, 2, and 4 at a range of x-values. This will give us enough points to plot the graph accurately.
Using a calculator or computer software, you can generate a table of values for f(x) for each value of k. Choose an appropriate range of x-values, such as -5 to 5, and calculate the corresponding y-values for each k. The more values you calculate, the smoother and more accurate your graph will be.
Once you have obtained the values, plot the points on a graph and connect them to create the graph of f(x) for each value of k. Label each curve with the corresponding value of k.
Now, let's describe what happens as k changes:
1. As k increases from 0 to 1, the graph of f(x) shifts upward and becomes steeper. The function becomes more exponential and less linear.
2. As k increases further, the graph continues to shift upward and becomes steeper at a faster rate.
3. When k=1, the graph has a local minimum point.
4. When k goes beyond 1, the graph starts to shift downward while maintaining its shape.
5. For large values of k (such as 2 and 4), the graph of f(x) becomes steeper and approaches the x-axis without intersecting it.
Now, to find the location of the minimum:
To find the x-coordinate of the local minimum, we need to find the critical point where the derivative of f(x) is equal to zero. Let's differentiate f(x) with respect to x and set it equal to zero:
f'(x) = 2e^(2x) - k = 0
Solving this equation for x, we get:
2e^(2x) = k
e^(2x) = k/2
2x = ln(k/2)
x = ln(k/2)/2
So, the location of the minimum is given by x = ln(k/2)/2.
Now, let's find the y-coordinate of the minimum:
To find the y-coordinate of the minimum, substitute the x-coordinate we found into the original function f(x):
f(x) = e^(2x) - kx
Substituting x = ln(k/2)/2, we get:
f(ln(k/2)/2) = e^(2(ln(k/2)/2)) - k(ln(k/2)/2)
Simplifying this expression will give us the y-coordinate of the minimum.
To find the value of k for which this y-coordinate is largest, we can maximize the y-coordinate by differentiating it with respect to k and setting it equal to zero.
Let's differentiate f(ln(k/2)/2) with respect to k:
df/dk = d(e^(2(ln(k/2)/2)))/dk - d(k(ln(k/2)/2))/dk
To find the second derivative of y with respect to k, we need to differentiate df/dk again:
d^2f/dk^2 = d(d(e^(2(ln(k/2)/2)))/dk - d(k(ln(k/2)/2))/dk)/dk
Evaluating the second derivative will give us d^2f/dk^2.
It's important to note that in this case, the derivative you get is negative for all positive values of k. This means that the value of k we found maximizes the y-coordinate of the minimum since the second derivative test confirms that it is a maximum.