Solving One Math Problem: Find H'(x), Concavity, and Tangent at x=4

  • Thread starter ChaNgeD
  • Start date
In summary, the student is trying to find where a horizontal tangent exists on a graph, and if so, locating a maximum or minimum. If it is above the curve, the line is above the graph. If it is below the curve, the line is below the graph. The student also asks for help with problems 1-4.
  • #1
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Help With One Math Problem URGENT!

Im sorry i have no work to show for this etc...but i need in the next 10-20 minutes or i fail this test...This depends on whether i pass the class this semester. ! Long Story to lazy to type and not much time!


A function H exists where H(4)= -3
H'(x)= x^2 - 2 / x

a)Determine where a horizontal tangent exists on the graph H, and if so locate a maximum/minimums. ( u can't do this with a graph it won't count, must be done algebraicly.)
b)Determine on what interval the graph H performs a concave down.
c) Determine the equation of the line where the tangent is at x = 4 On the original graph.
d)Is the Tangent of the line in C) below or above the curve? Explain.
 
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  • #2
Why do you have no work to show? It seems to me that you're just too lazy to do it, and want someone else to pass the class for you! Not only is this against PF rules, but it's pretty darn cheeky! I'm not doing it for you-- show some work and you may get help.

(P.S: If you haven't read the forum rules yet, see the sticky at the top of each homework forum)
 
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  • #3
It sounds like it was the quiz... a 10-20 minute quiz.

Good call cristo in not going for it!
 
  • #4
Your semester grade depends upon this? The answer depends entirely upon the definition of derivative which was probably covered in the first week of class! Sounds like you deserve to fail.
 
  • #5
ChaNgeD said:
Im sorry i have no work to show for this etc...but i need in the next 10-20 minutes or i fail this test...This depends on whether i pass the class this semester. ! Long Story to lazy to type and not much time!


A function H exists where H(4)= -3
H'(x)= x^2 - 2 / x

a)Determine where a horizontal tangent exists on the graph H, and if so locate a maximum/minimums. ( u can't do this with a graph it won't count, must be done algebraicly.)
b)Determine on what interval the graph H performs a concave down.
c) Determine the equation of the line where the tangent is at x = 4 On the original graph.
d)Is the Tangent of the line in C) below or above the curve? Explain.


Alrighty!
H'(x) is the derivative of H(x), so we need to find the integral of H'(x), which is H(x).
As you well know, the integral is taken by pieces (which, in this case, are x^2 and -2/x). To get the integrals, you simply add one to the exponent (the thing after the ^) of the variable, and divide the product by the new exponent. So the integral, or antiderivative, of x^6 is x^7/7. The integral of 2x^7=2x^8/8=x^8/4. Finally, the integral of x^2=x^3/3. (so the integral of x^a=x^(a+1)/(a+1))
The integral of the second piece, -2/x, is not so simple ^_^. You'd think the derivative of -2x^(-1) would be 2 (firstly, because 1/x=x^(-1) and secondly because the derivative of 2 is zero, and the point of an antiderivative is so that the derivative of the answer is the subject being antiderived!)
Well, the law I said holds true, but for 1/x, log|x| is the derivative. So our answer is -2log|x|.
Anyways, H(x) is now found to be x^3/3-2log|x|+C. At least I hope it is; my calculus skills are a little rusty. Oh; the +C comes in because when one derives, it would disappear (it may not be there, you say? Well, C can be zero).

Now we need to find 'c'!

We know that H(4)=-3
so..
-3=(4)^3/3-2log|4|+C
I don't know log|4| off the top of my head..
But we do know that 4^3=64, and that 64/3=21.333333333, so (-3-21.33333), which equals -24.333333/-2, which happens to equal 12.1666666, equals log|4|+C. Therefore, C=12.166666-log|4|, and
H(x)=x^3/3-2log|x|+(12.166666-log|4|)
You'll notice I used a different number of 6's (and if I didn't I intended to), because the 6's go on forever (it's actually 2/3's.)

Now the problem solving starts.

And it's not that, like the above posts, I don't believe you; it's just that I'm not quite up to solving such a difficult problem when it was due ~8 months ago.

Email me (mszlmb@hotmail.com) and I'll probably get back to you. Hey, I can tutor you too if you like :-p
Hope you passed.. Sort of (wouldn't hurt to actually learn the materials the second time..)
 
  • #6


Ran into this problem only by chance, but what overkill of an answer up there! This is a 20-min. calc 1 quiz question. a) set H' =0 solve x b) solve H" <0 c) standard point-slope formula for tangent line d) check concavity at x=4; down-->t.l. above curve; up -->t.l. below. Hope we're not too late for your quiz dude. If you don't make it as a scientist, there is aways room for another overpaid CEO.
 
  • #7


gammamcc said:
Ran into this problem only by chance, but what overkill of an answer up there! This is a 20-min. calc 1 quiz question. a) set H' =0 solve x b) solve H" <0 c) standard point-slope formula for tangent line d) check concavity at x=4; down-->t.l. above curve; up -->t.l. below. Hope we're not too late for your quiz dude. If you don't make it as a scientist, there is aways room for another overpaid CEO.

1 - he's right
2 - LOL!
 

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