Let f(x)=e5x−kx, for k>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.
To sketch the graph of the function f(x) = e^(5x) - kx for various values of k, we can use a calculator or computer software. Here is a description of what happens as k changes:
1. k = 1/9:
As k is a small positive value, the term -kx will have a relatively small effect on the graph compared to the exponential term e^(5x). The graph of f(x) will be similar to the graph of e^(5x), but slightly shifted downwards.
2. k = 1/6:
As k increases, the linear term -kx becomes more dominant, affecting the shape of the graph further. The graph will be shifted even more downward and flattened compared to the graph for k = 1/9.
3. k = 1/3:
As k continues to increase, the effect of the linear term becomes more pronounced. The graph will be shifted downwards and flattened even more compared to previous graphs. The slope of the graph will be steeper since k is larger.
4. k = 1/2:
At k = 1/2, the linear term -kx becomes equal to the exponential term e^(5x) at x = 0. The graph will intersect the y-axis at the point (0,0), and then diverge away from the x-axis as x increases. This value of k will result in a point of inflection on the graph.
5. k = 1:
At k = 1, the linear term -kx will have an equal effect to the exponential term e^(5x). The graph will intersect the y-axis at (0,-1), and then change direction as it passes through the point (0,0). The graph becomes concave down if x < 0 and concave up if x > 0.
6. k = 2:
As k increases further, the linear term -kx becomes the dominant factor, resulting in a significant downward shift of the graph compared to the exponential term. The graph will intersect the y-axis at (0,-2) and exhibit a steeper slope.
7. k = 4:
With k = 4, the linear term -kx is much larger than the exponential term e^(5x). The graph will be sharply downward, intersecting the y-axis at (0,-4). The exponential term e^(5x) will have a minimal impact on the overall shape of the graph.
In summary, as k increases, the linear term -kx becomes more significant, causing the graph to shift downwards and become flatter. The slope of the graph also becomes steeper. Conversely, as k decreases, the graph will shift upwards and become steeper as well.
To sketch the graph of the function f(x) = e^(5x) - kx for different values of k, you can follow these steps:
1. Choose a set of x-values to evaluate the function. For simplicity, let's choose x-values from -2 to 2 with a step size of 0.1. This range should be sufficient to capture the behavior of the function.
2. Calculate the corresponding y-values for each x-value using the given function and the values of k.
3. Plot the points (x, f(x)) on a graph and connect them to form a smooth curve.
Now, let's go through the process for the given values of k = 1/9, 1/6, 1/3, 1/2, 1, 2, and 4.
1. For k = 1/9:
Evaluate f(x) = e^(5x) - (1/9)x for x = -2, -1.9, -1.8, ..., 2.
Calculate the corresponding y-values using a calculator or computer.
2. Plot the points on the graph and connect them to form a curve. The resulting graph will be labeled G1.
3. Repeat the process for the other values of k (1/6, 1/3, 1/2, 1, 2, and 4), and label the resulting graphs G2, G3, G4, G5, G6, and G7, respectively.
Now, let's discuss how the graph changes as k varies.
1. When k = 1/9, the curve G1 will be relatively steep and slowly increasing. As x increases, the exponential term dominates, causing the function to grow rapidly.
2. As k increases to 1/6, the curve G2 shows a less steep slope. The linear term starts to have a more significant impact, causing the function to rise at a slower rate.
3. For k = 1/3, the curve G3 becomes even less steep. The linear term becomes more influential, leading to a gentler increase in the function.
4. When k = 1/2, the curve G4 exhibits a nearly linear growth. The exponential term becomes less dominant, and the function appears almost linear.
5. As k increases from 1/2 to 1, the curve G5 shows a similar trend to G4. However, the rate of increase becomes steeper again as k approaches 1.
6. For k > 1, such as k = 2 or 4, the curve G6 and G7 exhibit even steeper slopes. The exponential term dominates over the linear term, causing the function to grow rapidly as x increases.
In summary, as k increases, the impact of the linear term in the function becomes more pronounced, leading to a slower growth rate. Conversely, as k decreases, the exponential term dominates, resulting in a steeper increase in the function.
Can't do it until you show what e5x means
is it:
e^(5x) or (e^5)(x) or e(5x) or ...
assuming the usual carelessness with parentheses, I'll go with
y = e^(5x)-kx
y' = 5e^(5x) - k
y" = 25e^(5x)
so, now we can answer the questions
y'=0 at x = 1/5 ln(k/5)
y" is always positive, so
there is a minimum at (1/5 ln(k/5) , k/5 (1-ln(k/5))
Now, which value of k makes y=k/5 (1-ln(k/5)) a maximum is k=5
since
y' = -1/5 ln(k/5)
y" = -1/5k < 0 at k=5