SKetching Concavity, Inflection Points, and Intervals of Increase and Decrease

[ardentmed]

Hey guys,

I have a couple of questions about this problem set I've been working on. I'm doubting some of my answers and I'd appreciate some help.

Question:

For \(\displaystyle f(x)\, =\, \dfrac{x^2}{(x\, -\, 2)^2},\) give the intervals of increase and decrease, identify any local extrema, give the intervals where concave up and concave down, and the coordinates of any points of inflection. Identify any asymptotes and sketch the curve.

For \(\displaystyle f(x)\, =\, 3x^{\frac{2}{3}}\, -\, x,\) give the intervals of increase and decrease, identify any local extrema, give the intervals where concave up and concave down, and the coordinates of any points of inflection. Identify any asymptotes and sketch the curve.

For 1, I simply calculated f'(x)=0 and f'(x) = DNE. Which gave me f(0)=0 and x=2 as the critical points. Then when determining concavity via f''(x)=0 and f''(x)= DNE, I got f''(x)=DNE at x=2 (again). How would I go about assessing this question? What feature is x=2 supposed to represent. Moreover, should I confirm my answers with the first and second derivative tests, or is there a different test I should be using?

As for 2, for f ' (x), I got:
f(2^(1/3)) = 4.74.

But I was puzzled when I found out that no inflection point exists when I assessed f''(x)=0. Did I miscalculate anything up to this point?

Thanks in advance.

 

[Quaid]

f(0)=0 and x=2 as the critical points

This is correct, but the statement ought to read, "x=0 and x=2 as the critical points".

when determining concavity via f''(x)=0 and f''(x)= DNE, I got f''(x)=DNE at x=2 (again).

You are correct that the second derivative does not exist at x=2.

What is your solution for f''(x) = 0 ?

How would I go about assessing this question?

The critical points x=0 and x=2 divide the Real number line into three intervals.

Check the sign of the first derivative, in each of these intervals, by evaluating f'(x) at a test value for x.

Function f increases in intervals where its first derivative is positive; function f decreases in intervals where its first derivative is negative.

Please show us your answers for where f is increasing and decreasing.

Then, answer the question highlighted in red above, and we can continue from there.

What feature is x=2 supposed to represent.

An asymptote.

should I confirm my answers with the first and second derivative tests

You need to generate answers, before you can confirm them. Use these tests to generate your answers.

---

As for 2, for f ' (x), I got:

f(2^(1/3)) = 4.74

f'(x) is not a number.

Please start a new thread for the second exercise, and show your work on the first derivative.

Both of your functions in this thread are named f. It's confusing, to discuss them both at once. This is why the forum guidelines request separate threads for separate exercises.

Here is a link to the summary page of the forum guidelines.

Cheers

 

[Quaid]

Hi ardentmed,

Are you still stuck?

Do you need more help, finding the intervals where f is increasing/decreasing?

Cheers

 

[ardentmed]

Hi ardentmed,

Are you still stuck?

Do you need more help, finding the intervals where f is increasing/decreasing?

Cheers

That part should be relatively easy, but I'm doubting my intervals for concavity for both but moreso for the second one.

Thanks again for all the help. I greatly appreciate it.

 

[Quaid]

[Finding intervals of function increase and decrease] should be relatively easy

Agree!

Post your work and results for this, on the first exercise, if you'd like us to check 'em.

I'm doubting my intervals for concavity for both but moreso for the second one

For the first exercise, did you solve f''(x) = 0 ?

That is, what intervals did you check for concavity?


Please start a new thread, for the second exercise. Include all of your work and reasoning, so far, and we can go from there.

Cheers

 

[ardentmed]

Here is the result for the first graph. Am I on the right track?

Inflection points were found at x=-1 and a critical point (minimum) was found at x=0.

 

[ardentmed]

I should also state that there is no symmetry since f(x) =/ (f(-x).

Since there is an increasing slope to the left of f(0) and a decreasing slope to the right thereof, there must be a local minimum at f(0) = 0.

Furthermore, there is an inflection point at f(-1) = 1/9 and f''(x) = DNE at x=2 (and the same applies for f'(x)).

Because the limit as x-> infinity is 1, there is a horizontal asymptote at y=1. Moreover, a vertical asymptote exists at x=2 since taking the limit as x -> 2 yields an undefined value.

As such, f increases from (0,2) and decreases from (-infinity, 0) u ( 2,infinity).

Moreover, f is concave up from (-1,infinity) and concave down from (-infinity, -1).

Intercepts are both at (0,0) for the x and y-intercepts respectively.

Am I on the right track? Thanks again for the help, guys. I really appreciate it.