napierian logarithm problem

[red and white kop!]

the function f is defined for positive real values of x by f(x)=12lnx - x^(3/2)
the curve crosses the x-axis at points A and B
a. by calculation, show that the value of x at point A lies between 1.1 and 1.2
b. the value of x at the point B lies in the interval (n, n+1) where n is an integer. find n

for a. i took 12lnx = x^(3/2) and turned it all around, getting e^x^(3/2) = x^12 etc. but i still have no idea of how to get the interval for x
for b. i have no idea

 

[BigGlenntheHeavy]

\(\displaystyle f(x) \ = \ 12ln|x|-x^{3/2}, \ Note \ f(x) \ is \ continuous \ on \ [1.1,1.2].\)

\(\displaystyle f(1.1) \ < \ 0 \ and \ f(1.2) \ > \ 0, \ hence \ by \ the \ IVT \ there \ exist \ an \ f(c) \ = \ 0.\)


\(\displaystyle b. \ f(8) \ > \ 0, \ f(9) \ < \ 0, \ hence \ n \ = \ 8\)

 

[red and white kop!]

im not sure i understand you; cant one get this by working through the equation?

 

[BigGlenntheHeavy]

Observe the graph of f(x). f(x) crosses the x axis (y=0) at 1.1010683 and 8.797409.

By calculation Newton's method will give you these values as it is impossible to solve f(x) by ordinary algebraic means.


[attachment=0:cy2k04dg]qqq.jpg[/attachment:cy2k04dg]

 

[Deleted member 4993]

im not sure i understand you; cant one get this by working through the equation?

Yes you can - and that is how Glen provided the answer. Study his first response carefully.

Tell - why do you think that - anything but the algebraic evaluation of the function has been used there (except of course use of a theorem).

Have you not studied the Intermediate Value theorm? It is actually quite intuitive.