Calculating intercept with x-axis

  • Thread starter GregA
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I am looking more for a better method of solving this exam question as opposed to a solution (already solved but in one area I believe, badly.)
f(x) =12lnx - x^(3/2)
the graph cuts the x-axis at A reaches a maximum point at C then cuts again at B.
A) show by calculation that the x coordinate at A lies between 1.1 and 1.2
b) b lies in the interval (n,n+1) where n is an integer. Determine B
c) find the value of x for which dy/dx = 0 and hence find the maximum value of C
d) find the range of values of x for which f(x) is increasing
c) and d) are pretty straight forward...
the only way I could calculate a) was to start by saying...lnx>1 (as with a lower value... x^(3/2) would be less than 1..lnx for x<1 yields a negative result) using a value of 1 for x I found that lnx = 1^(3/2)/12 and this gives a value of 1.087...x must now be greater than this number, repeating this method I found x around 1.101068...
question b) is my problem... how can the x position at B be calculated? the above method fails. (I only solved it by throwing a few numbers at it until I hit the correct range...I know from question c that I'm looking for a value greater than 4)
my knowledge so far is limited to differentiating and integrating polynomials and lnx, finding areas and stationery values etc...I haven't covered much more so far and have certainly not done any work on soving these types of problems...there must surely be a solution however.
 
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  • #2
HallsofIvy
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GregA said:
I am looking more for a better method of solving this exam question as opposed to a solution (already solved but in one area I believe, badly.)
f(x) =12lnx - x^(3/2)
the graph cuts the x-axis at A reaches a maximum point at C then cuts again at B.
A) show by calculation that the x coordinate at A lies between 1.1 and 1.2
b) b lies in the interval (n,n+1) where n is an integer. Determine B
c) find the value of x for which dy/dx = 0 and hence find the maximum value of C
d) find the range of values of x for which f(x) is increasing
c) and d) are pretty straight forward...
the only way I could calculate a) was to start by saying...lnx>1 (as with a lower value... x^(3/2) would be less than 1) using a value of 1 for x I found that lnx = 1^(3/2)/12 and this gives a value of 1.087...x must now be greater than this number, repeating this method I found x around 1.101068..
The problem does not ask to find where the x-intercept is- only to show that there must be one between 1.1 and 1.2. What is f(1.1)? What is f(1.2)? What does that tell you?

question b) is my problem... how can the x position at B be calculated? the above method fails. (I only solved it by throwing a few numbers at it until I hit the correct range)
The problem said "b) b lies in the interval (n,n+1) where n is an integer. Determine B" Once again, it does not ask for a precise value.
Find f(2), f(3), f(4), etc. until the sign changes.
("Trial and error" is a perfectly valid mathematical method!)
 
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Use the http://mathworld.wolfram.com/Mean-ValueTheorem.html" [Broken] on a)
 
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  • #4
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Thanks for the replies folks...when the question asked me to *show by calculation* that x lies between 1.1 and 1.2 HallsofIvy, I did actually try f(1.1) and f(1.2) but was under the impression that being a question at the tail end of 48 questions, being able to find accurate values for x was expected...( it just annoys me when I believe there's an answer somewhere that I cannot find :mad: )
Thanks for the link Incredible...There is no reference to that theorem in the textbook I'm using right now but I will certainly have a good look at it. :smile:
 

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