How Can I Solve These Math Problems for My Exam Tomorrow?

  • Thread starter Thread starter GrP
  • Start date Start date
  • Tags Tags
    Exam
Click For Summary
SUMMARY

This discussion focuses on solving two specific math problems: finding stationary points of the function f(x) = (x^4)^(x^2) and evaluating the improper integral from 1 to infinity of the function ((x^2+1)^(1/2))/((x^3)*(x^2-1)^(1/2)). For the first problem, participants confirm the use of logarithmic differentiation and the second derivative test to identify local and global extrema. For the integral, u-substitution with u = x^2 + 1 is recommended, along with potential trigonometric substitution for simplification.

PREREQUISITES
  • Understanding of logarithmic differentiation
  • Familiarity with the second derivative test for extrema
  • Knowledge of improper integrals
  • Experience with u-substitution in integration
NEXT STEPS
  • Study logarithmic differentiation techniques
  • Learn about the second derivative test for identifying local minima and maxima
  • Research methods for evaluating improper integrals
  • Explore u-substitution and trigonometric substitution in integral calculus
USEFUL FOR

Students preparing for math exams, particularly those focusing on calculus concepts such as differentiation and integration techniques.

GrP
Messages
1
Reaction score
0
excuse me for my bad english, but I'm trying hard.

1. function f(x), for x>0 is defined with:

f(x)=(x^4)^(x^2)

^ - potency

find stationary points of the function f(x) on the interval (0,infinite).

what exactly do i do here?
do i use logarithm on both sides, and then i get:

log(y)=(x^2)*log(x^4)

and then i different (differential) both sides, and set y'=0
and i calculate x?
is this right so far?

now i want to get local min. and max.
and i different again, and then what?
and i need to find out if these points are globlal min and max.


2. integral from 1 to infinity (((x^2+1)^(1/2))/((x^3)*(x^2-1)^(1/2)))dx

and ideas what could i do? what would "u" be? (u^2)=(x^2)+1?

please help me asap
thank you!
 
Physics news on Phys.org
1. Thats right so far.
2.

[tex]\int_1^{\infty} \frac{\sqrt{x^2+1}}{x^3\sqrt{x^2-1}}dx[/tex]

Well first off, you'll need to do an improper integral.

First thing I would try is a trig substitution, then maybe look at it by parts if that doesn't work. u = x^2 might get you somewhere but I don't know.
 


Hi there, it seems like you are struggling with some math problems for your exam tomorrow. Don't worry, I will try my best to help you out.

For the first problem, finding the stationary points of the function f(x), you are on the right track. Yes, you can take the logarithm on both sides and then differentiate both sides to find the critical points. But before that, you need to simplify the expression first. Starting with f(x)=(x^4)^(x^2), you can rewrite it as (x^2)^(2x^2), and then take the natural logarithm on both sides to get ln(f(x))=2x^2ln(x^2). From here, you can differentiate both sides and set the derivative equal to 0 to find the critical points. Once you have the critical points, you can then use the second derivative test to determine if they are local minima or maxima. And to check if they are global minima or maxima, you can use the first derivative test or compare the values of f(x) at the critical points.

For the second problem, finding the integral from 1 to infinity, you can try using u-substitution. Let u=x^2+1, then du=2xdx. You can substitute this into the integral and then try to simplify it from there. As for the second part of the question, u would be equal to (x^2)+1, not (u^2)=(x^2)+1. Remember, when you are using u-substitution, u should be a function of x, not the other way around.

I hope this helps you with your math problems. Remember to stay calm and focused during your exam tomorrow and try your best. Good luck!
 

Similar threads

  • · Replies 18 ·
Replies
18
Views
2K
Replies
14
Views
5K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
8
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
Replies
2
Views
1K
Replies
3
Views
993
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
Replies
0
Views
2K