Question

The function f has first derivative given by f'(x)=(x)^1/2/1+x+x^3. What is the x-coordinate of the inflection point of the graph of f?
a. 1.008
b. 0.473
c. 0
d. -0.278
e. the graph has no inflection points

Answer 1

Find the numbers not in the domain of, f'(x)
Set the f'(x) = 0, and solve for x
Create intervals out the numbers generated and input numbers inbetween these intervals.

If you get a positive number out
f'(x)>0, thus the function is increasing
if you get a negative number out
f'(x)<0, thus the function is decreasing

an inflection point is likely when you have two intervals where the function is increasing or decreasing.

They look like this

Answer 2

Note that the intervals have to be next to eachother.

Answer 3

could you show thiswork please?

Answer 4

for example a functin such as f'(x) = 3x/(x-1)
we see the domain is All real numbers - (+1)
so we have the number 1,

we set f'(x) = 0

0(x-1) = 3x(x-1)/(x-1)
0 = 3x
x = 0
we have a second number 0

so we have two numbers to generate intervals for 0, 1

we have the interval (-infinity, 0), (0,1), (1, infinity)
we then make a table



Now insert numbers inbetween each interval into f'(x)

so we will check (-infinity,0), lets use -1 as it belongs to this interval

f'(-1) = 3(-1)/((-1-1)
f'(-1) = -3/-2 = 3/2 Thus we have a positive the interval is increasing We know this because the rules:

f'(x)>0, thus the function is increasing
f'(x)<0, thus the function is decreasing

Thus we have that