Sketching graphs with extreme values with the Given Information

[ardentmed]

Hey guys,

I'm having trouble with this problem set I'm working on at the moment. I'd appreciate some help with this question:

(I'm only asking about question one. Please ignore question two)

I'm having trouble sketching this graph out. If f' and f'' are not defined at 2, does that mean that they are infinite? Also, is the limit as x approaches infinite is infinite, does that mean that it is pointing in the upwards direction and that there is no horizontal asymptote? Moreover, if f'(5)=0, I'm assuming that that means there is some sort of critical point present, in which case it should be a maximum according to the increasing and decreasing values surrounding x=0.

Am I on the right track? Can someone give me an estimation of what the graph should look like? I can upload a sketch of mine (which is probably partially incorrect), but a clean computer sketch would do wonders for me to visualize the problem.

Thanks in advance.

 

[Dethrone]

Let's see, if f' and f'' are not defined at 2, they could be any sort of discontinuity, such as a removable discontinuity, but more commonly a vertical asymptote if it is a rational function.
\(\displaystyle
\lim_{{x}\to{\infty}} f(x) = \infty \) implies that as x approaches infinity, your graph will go in the positive infinity direction. So yes, it will point upwards like a parabola, and it will not have a horizontal asymptote. It will only have one if it approaches a finite value as \(\displaystyle \lim_{{x}\to{\pm\infty}}\).

The problem tells us that as you approach 5 from the negative side , the slope is negative and as you approach 5 from the positive side, the slope is positive. If the slope goes from negative to positive, it must be a minimum. In terms of increasing and decreasing, a negative slope implies decreasing, and a positive slope implies increasing. Therefore, going from decreasing to increasing is characteristics of a minimum. With that said, if you provide a picture of your sketch, I could pinpoint the errors, if there are any.

 

[ardentmed]

View attachment 2905

Does that look right?

Thanks.

Let's see, if f' and f'' are not defined at 2, they could be any sort of discontinuity, such as a removable discontinuity, but more commonly a vertical asymptote if it is a rational function.
\(\displaystyle
\lim_{{x}\to{\infty}} f(x) = \infty \) implies that as x approaches infinity, your graph will go in the positive infinity direction. So yes, it will point upwards like a parabola, and it will not have a horizontal asymptote. It will only have one if it approaches a finite value as \(\displaystyle \lim_{{x}\to{\pm\infty}}\).

The problem tells us that as you approach 5 from the negative side , the slope is negative and as you approach 5 from the positive side, the slope is positive. If the slope goes from negative to positive, it must be a minimum. In terms of increasing and decreasing, a negative slope implies decreasing, and a positive slope implies increasing. Therefore, going from decreasing to increasing is characteristics of a minimum. With that said, if you provide a picture of your sketch, I could pinpoint the errors, if there are any.

 

[Dethrone]

By inspection, your limits to infinity are wrong. Your graph should point upwards to infinity as x goes to infinity. The first derivative is greater than 0 when x < 2, or x > 5, meaning that your graph should be increasing in said intervals, which I do not see reflected in your graph. Posting a picture of your graph was very helpful, as I could see where your mistakes were.

 

[ardentmed]

By inspection, your limits to infinity are wrong. Your graph should point upwards to infinity as x goes to infinity. The first derivative is greater than 0 when x < 2, or x > 5, meaning that your graph should be increasing in said intervals, which I do not see reflected in your graph. Posting a picture of your graph was very helpful, as I could see where your mistakes were.

Bear in mind that the webcam took the photo in a way that mirrored the x axis. Is it still incorrect?

Thanks again.

Edit: Oh, so it should be pointed upwards in both instances? Isn't x -> -infinity when x<2?

 

[Dethrone]

No, just in the instance when x goes to positive infinity. I will recheck your graph again, I did not know that it was reflected.

- - - Updated - - -

It looks right now. (Yes)

 

[Dethrone]

You figured it out, but I also should have added that $f'$ and $f''$ being not defined could also imply some sort of kink, corner, cusp, etc on the graph. These situations are not defined because the derivative is different as you approach from the left and right sides:

$$\lim_{{h}\to{0^+}}\frac{f(x+h)-f(x)}{h} \ne \lim_{{h}\to{0^-}}\frac{f(x+h)-f(x)}{h}$$