Graph a function

Question

Graph of a function with the given listed below

continuous for all real numbers x

first derivative

f" (x)>0 if |x| <2 and f'(x) <0 if |x| >2

f" (2)=0

lim x approaches infinity f(x)=1

f is an odd function

second derivative

f" (x) <0 if 0<x<3 and f"(x) >0 if x >3

Answer 1

Answer 2

Increasing from x=-2 to x=2

Goes through origin

Concave down from x=0 to x=3

Concave up from x=0 to x=-3

Max at x=2

Min at x=-2

Horizontal asymptote at y = 1 Because f approaches 1 as x approaches + infinity. (beyond x=3)

Horizontal asymptote at y = -1 Because f approaches -1 as x approaches - infinity (beyond x=-3)

Sorry, don't know how to upload pictures.

Answer 3

Let's take a look at each condition to see how it affects the final graph:

continuous for all real numbers x: This means that there are no holes, jumps, or vertical asymptotes and a y-value exists for every x-value.

lim x approaches infinity f(x)=1: This means that there is a horizontal asymptote at y=1.

f is an odd function: This means that f(-x) = -f(x) and the function is symmetric about the origin.

As far as the first and second derivatives, use the information they give you to make a chart of how the function acts between points. The critical points are when the first derivative changes from positive to negative and indicate where the graph of f(x) will change from increasing to decreasing. At the critical points, the first derivate will be 0, which tells us that the slope of f(x) is 0 and is horizontal at that point. This will usually either create a local minimum or maximum. The inflection points are when the second derivative changes from positive to negative and indicate where the graph of f(x) will change concavity. Follow the rules below:

f'(x) > 0: The first derivative is positive and f(x) is increasing

f'(x) < 0: The first derivative is negative and f(x) is decreasing

f'(x) = 0: The slope of f(x) is 0 which indicates a local minimum or maximum

f"(x) > 0: The second derivative is positive and f(x) is concave up

f"(x) < 0: The second derivative is negative and f(x) is concave down

Applying the rules to your problem, make a chart that includes all critical and inflection points and note the behavior of f(x) between each point. Critical points occur at x=-2,2. Inflection points occur at x=-3,0,3. Then graph the critical and inflection points and make a curve between the points that follows the behavior. Once you draw the graph from 0 < x < infinity, it will be easier to determine the behavior from -infinity < x < 0 based on the fact that the function is odd and symmetric about the origin.

-infinity < x < -3: f'(x) < 0 so f(x) is decreasing, f"(x) < 0 so f(x) is concave down

-3 < x < -2: f'(x) < 0 so f(x) is decreasing, f"(x) > 0 so f(x) is concave up

-2 < x < 0: f'(x) > 0 so f(x) is increasing, f"(x) > 0 so f(x) is concave up

0 < x < 2: f'(x) > 0 so f(x) is increasing, f"(x) < 0 so f(x) is concave down

2 < x < 3: f'(x) < 0 so f(x) is decreasing, f"(x) < 0 so f(x) is concave down

3 < x < infinity: f'(x) < 0 so f(x) is decreasing, f"(x) > 0 so f(x) is concave up