Number of results: 33,438
Find the relative extrema of the functions. f(x)=4x/(x^2+1) f(x)=1/(x-2) take the derivative, set to zero. the first as an example: f(x)= 4x (x^2 +1)^-1 f'(x)= 4 (x^2 +1)^-1 -4x(2x)(x^2 +1)^-2 set to zero. Multiply thru by (x^2+1)^2 0=4(x^2+1) - 8x^2 4x^2=1 x= +- 1/2 what ...
Wednesday, February 28, 2007 at 5:17pm by Reisha
relative extrema x^4-2x^2+5 so far I know how to find the derivative which is 4x^3-2x now I am stuck... Please help Huh? 4x^3-2x=0 2x(2x^2-1)=0 and what are the roots? I will give you a hint: one is x=0 The other two are in the second parenthesis. Is it x-1 and x=-1 I did a ...
Friday, March 9, 2007 at 10:51am by dee
relative extrema x^3 + 1/2x^2 - 2x + 5 Ok I have 3x^2 + x - 2 = 0 Am I on the right track? I think i know why I kept on asking for help with these pribelms. I forgot the step of factoring. Please check... 3x^2 + x - 2 = 0 (3x-2)(x+1)=0 x=2/3, x=-1 Correct. Nice job. If you ...
Friday, March 9, 2007 at 11:14am by dee
case 1. Find the absolute maxima and minina values of f(x)=sin2x+cos2x on [0, pie] Specify both the x and y coordinateof the absolute maximum and absolute minimum. Gives answers in exact form. case 2. Use the second derivative test for relative maxima nd minima to find the ...
Sunday, June 24, 2007 at 4:07pm by Mae
I have the equation: cos x - x = f(x) I am told to find the relative extrema. I am also told to use the Second Derivative Test where applicable. My question is how do I solve this problem, and how do I know when to use the Second Deriv test? Thanks!
Wednesday, November 14, 2007 at 12:13am by Phil
Marlyn: I removed your someone post. What drwls is saying, it appears that you are answer grazing, you show now work on a variety of topics, not indicating what you dont understand about the question. Surely you know something..that is our starting point for teaching. We cant ...
Sunday, December 6, 2009 at 5:41pm by bobpursley
Find the relative extrema of function. f(x,y)=x2+3xy+y2-10x-20y+12
Tuesday, October 19, 2010 at 3:22am by AYA
Find the relative extrema of f(x) = -2x^2 + 4x + 3 I don't know how to use the First derivative test to find it.
Sunday, October 31, 2010 at 9:30pm by jack
Find all relative extrema and points of inflection of the function: f(x) = sin (x/2) 0 =< x =< 4pi =< is supposed to be less than or equal to. I can find the extrema, but the points of inflection has me stumped. The inflection point is (2pi,0) but shouldn't there be ...
Sunday, November 14, 2010 at 11:48pm by James
If g is a differentiable function such that g(x) is less than 0 for all real numbers x and if f'(x)=(x2-4)g(x), which of the following is true? f has a relative maximum at x=-2 and a relative minimum at x=2, f has a relative minimum at x=-2 and has a relative maximum at x=2, f...
Sunday, January 2, 2011 at 9:06pm by felix
Find the intercept, relative extrema, point of inflection, and asymptotes of the function(if they exist)of: y = (x^2 + 1)/(x^2 - 9)
Monday, January 10, 2011 at 4:39pm by Jands
Find the intercept, relative extrema, point of inflection, and asymptotes of the function(if they exist)of: y = (x^3)/(x^2 - 4)
Monday, January 10, 2011 at 4:41pm by Jands
Find the intercept, relative extrema, point of inflection, and asymptotes of the function(if they exist)of: y= 3[(x-1)^(2/3)] - [(x-1)^2]
Monday, January 10, 2011 at 4:43pm by Jands
Find the intercept, relative extrema, point of inflection, and asymptotes of the function(if they exist)of: y= (x^4) - 8(x^3) + 18(x^2) - 16x + 5
Monday, January 10, 2011 at 4:46pm by Jands
The intercept is the y value when x = 0. That's pretty easy. The relative extrema occur where dy/dx = 0 The inflection point is where d^2y/dx^2 = 0 The asymptotes are where the denominator of y(x) is zero: x = 3 and -3. Now you have some differentiating to do.
Monday, January 10, 2011 at 4:39pm by drwls
I'm doing test corrections, and it's been a few weeks since we did this, so I'm a little fuzzy. I'm supposed to find the zeros, relative extrema, and inflection points of f(x) = x^3 - 6x^2 + 3x + 10. The zeros would be -1, 2, and 5 right? Then I would take the derivative of ...
Tuesday, December 13, 2011 at 12:01pm by Sarah
Find all the relative extrema of f(x) = x = x3 – 3x2 + 4.
Wednesday, January 18, 2012 at 3:44pm by Jordan
Basic Calculus-Relative Extrema
Hello, please help me :) A question asked that I find all relative extrema of a function f(x)= x + 1/x. I'm a bit confused with my solution, but I'll show my work first..: I found f'(x) to get the critical numbers: f'(x)=1-x^-2 f'(x)=(x^2 -1)/x^2 f'(x)=(x+1)(x-1)/x^2 therefore...
Thursday, March 15, 2012 at 11:29pm by mary
Basic Calculus-Optimization Problems
Hello, please help me. There is a question where I have to optimize the area of a field that is being fence in like this: [|] i.e. two rectangle fields, side by side: area= 2x by y. I get to use 200ft of fencing. From that diagram and the amount of fencing I have, I made this ...
Friday, March 16, 2012 at 1:55am by mary
Find any relative extrema and any points of inflection if they exist of f(x)=x^2+ ln x^2 showing calculus. Please show work in detail so I can follow. Thanks. The answer is no relative max or min and the points of inflection are (1,1) and (-1,1) but I'm having trouble getting ...
Saturday, May 4, 2013 at 1:12pm by Rene