Quick Calculus 1 Homework Help Needed: Final Exam Prep

[thercias]

Homework Statement

The Attempt at a Solution

I figured out #11, could use help with the other questions. I have a final tomorrow in the morning and my professor gave us a few questions to try out so help would be appreciated. For #8 I'm not sure how to draw the second derivative graph, for #12 I don't know where to begin, and for #15 I don't know how to find the anti derivative of f'(x) so I can't start the problem. Help is appreciated

 

[Magnawolf]

For 8 to draw f'(x) you need to see where the slopes are increasing, decreasing or remaining constant. From 0 to 3, you can see the slope is constant. From 3 to 6-7ish the slopes are increasing and from 7 on the slope is decreasing. You can draw a graph based on the knowledge I just told you. Do the same to that graph to get the 2nd derivative graph.

 

[thercias]

That's what I pretty much thought, thanks for the reasurement. The only one I really don't got a clue on is #12, so if someone could explain that for me that'd be great.

 

[Mentallic]

For 8) once you draw the derivative f '(x) then just draw the derivative of that graph to find f ''(x).


For 12) remember that you can split an integral up into sections, for example,
\int _ 1 ^5 f(x)dx = \int _ 1 ^2 f(x)dx + \int _2 ^ 4 f(x)dx + \int _4 ^5 f(x)dx
so you should do the same, choosing your sections based on the changing slope of the graph of f.


For 15) you aren't expected to take the anti-derivative. This should be obvious because it's giving you a lot more information than just the derivative function.
So to start drawing the function, begin with points that you can quickly determine, such as f(0) = 1. Once you draw this point on your axis, also plot the limits at the ends of your graph.

Now, what can we determine from the derivative and second derivative? Well, f ''(x) has a perfect square in the denominator so it's always positive (except at x=\pm 2) and so since the numerator is -8x, this means that for positive x values, the second derivative is negative, and for negative values it's positive. This tells you what the concavity of the graph is (think of the shape of a parabola x² that has positive concavity versus a parabola -x² which has negative concavity).

What about the first derivative? We already know there's a problem at x=\pm 2 because the denominator is 0 at those points, so what is

\lim_{x\to 2^-} f'(x)

and

\lim_{x\to 2^+} f'(x)

and the same for -2?

 

[thercias]

would the answer to this just be 0.5(4+6+4)=7?

 

[thercias]

For 8) once you draw the derivative f '(x) then just draw the derivative of that graph to find f ''(x).


For 12) remember that you can split an integral up into sections, for example,
\int _ 1 ^5 f(x)dx = \int _ 1 ^2 f(x)dx + \int _2 ^ 4 f(x)dx + \int _4 ^5 f(x)dx
so you should do the same, choosing your sections based on the changing slope of the graph of f.


For 15) you aren't expected to take the anti-derivative. This should be obvious because it's giving you a lot more information than just the derivative function.
So to start drawing the function, begin with points that you can quickly determine, such as f(0) = 1. Once you draw this point on your axis, also plot the limits at the ends of your graph.

Now, what can we determine from the derivative and second derivative? Well, f ''(x) has a perfect square in the denominator so it's always positive (except at x=\pm 2) and so since the numerator is -8x, this means that for positive x values, the second derivative is negative, and for negative values it's positive. This tells you what the concavity of the graph is (think of the shape of a parabola x² that has positive concavity versus a parabola -x² which has negative concavity).

What about the first derivative? We already know there's a problem at x=\pm 2 because the denominator is 0 at those points, so what is

\lim_{x\to 2^-} f'(x)

and

\lim_{x\to 2^+} f'(x)

and the same for -2?

So for #12, the answer for g(1) would be -1 because the slope is -1, the anti derivative of that is -x and just plug in 1 and you get the answer? Also how would you find the answer for something like g(3) where the point is between the changing slopes.

I'm kind of lost on #15 now, will give a try though. Thanks for the help.

 

[SammyS]

So for #12, the answer for g(1) would be -1 because the slope is -1, the anti derivative of that is -x and just plug in 1 and you get the answer? Also how would you find the answer for something like g(3) where the point is between the changing slopes.

No. g(1) ≠ -1 .

If $\displaystyle \ g(x)=\int_{1}^{x}f(x)\,dx\,,\$ then $\displaystyle \ g(1)=\int_{1}^{1}f(x)\,dx\ .\$

Think of the definite integral as the area under the curve -- or the negative of the area above the curve.

 

[thercias]

No. g(1) ≠ -1 .

If $\displaystyle \ g(x)=\int_{1}^{x}f(x)\,dx\,,\$ then $\displaystyle \ g(1)=\int_{1}^{1}f(x)\,dx\ .\$

Think of the definite integral as the area under the curve -- or the negative of the area above the curve.

ok i think i got it. so i don't really even need to take the antiderivative..just look at the graph and calculate area. g(1) = 0