How derivatives affect graphs homework help

In summary, the problem is to find the intervals of increasing and decreasing, relative extrema, intervals of concavity, and inflection points for the given function. The first derivative is used to find the intervals of increasing and decreasing, while the second derivative is used to find the intervals of concavity and inflection points. The critical points for the first and second derivative are x = 0 and x = 1, and x = 0 and x = 4 respectively. By plugging in these values, the intervals and relative extrema can be determined.
  • #1
h20skier
1
0
Hello All, This problem comes from #18 of section 4.3 out of stewarts 6th edition.

The Question:

Consider the following function:
f(x)= e^(-x) * sqrt(x)

What I am trying to find.

a.) Find the intervals on which f is increasing and decreasing.
b.) find any relative extrema.
c.) find the intervals of concavity.
d.) find all inflection points.

For part a. I have taken the first derivative and got. 1/2x^-(1/2) *e^-x - e^-x * sqrt (x)

after simplifying i got e^-x= 0
and 1/2 x^-1/2 - sqrt x = 0

I know i need to find these for my critical points but I am having difficulty proceeding.

for extrema i know i plug in values between my critical points to see the critcal values.

after that I know I take the 2nd derivative to find critical points for my concavity.

thanks for your help!
 
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  • #2
For part a: The first derivative of the given function is f'(x) = e^(-x) * (1/2)*x^(-1/2) - e^(-x) * sqrt(x). The critical points occur when f'(x) = 0. Solving for x, you get x = 0 and x = 1. Therefore the intervals on which f is increasing are (0,1) and decreasing are (-∞,0] and [1, ∞). For part b: To find any relative extrema, you need to plug in the critical points and the endpoints of the interval into the original function to see if they are maxima or minima.At x = 0, f(0) = 0. At x = 1, f(1) = e^(-1) * sqrt(1) = 1/e. Therefore, the relative minimum is at x = 0 and the relative maximum is at x = 1. For part c: The second derivative of the given function is f''(x) = e^(-x) * (-1/4)*x^(-3/2) - (1/2)*e^(-x) * x^(-1/2) - e^(-x) * (1/2)*x^(-1/2). Setting f''(x) = 0, you get x = 0 and x = 4. Therefore, the intervals on which f is concave up are (0,4) and concave down are (-∞,0] and [4,∞). For part d: The inflection points occur when the concavity of f changes. Using the intervals found in part c., the inflection points occur when x=0 and x=4.
 

FAQ: How derivatives affect graphs homework help

How do derivatives affect the slope of a graph?

Derivatives represent the rate of change of a function at a given point. Therefore, the derivative affects the slope of a graph by determining the steepness or direction of the line at that specific point.

Can derivatives change the shape of a graph?

Yes, derivatives can change the shape of a graph by altering the curvature or direction of the function. For example, a positive derivative will result in a concave up graph, while a negative derivative will result in a concave down graph.

How do derivatives affect the maximum and minimum points of a graph?

The derivative of a function can help determine the maximum and minimum points on a graph. At these points, the derivative will be equal to zero, indicating a change in direction of the slope. This can help identify the location of local extrema.

How can derivatives help in graphing complicated functions?

Derivatives can help in graphing complicated functions by providing information about the behavior of the function at different points. This can help determine the shape, slope, and concavity of the graph, making it easier to accurately plot the function.

Can derivatives be used to find the rate of change of a graph?

Yes, derivatives can be used to find the rate of change of a graph. This is because the derivative represents the instantaneous rate of change of a function at a specific point. By evaluating the derivative at different points, the overall rate of change of the function can be determined.

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