SOLUTION: If f(x)= (k+x³sec(x))/(sin(b)×(x-π)) is continuous at x= π, and kb -f(π)= 4(π^(5/ 2) , find the value of b .

sec(x) = .


When  x  approaches to π,  then cos(x)  approaches to -1; hence, sec(x) approaches to -1.

It means that  k+x^3*sec(x)  in the numerator is close to  k-x^3.


In the denominator, sin(b) is a non-zero constant, so, we should not worry about it.
But in the denominator, we have x-π, and, in order for the expression be a continuous function, 
k-x^3 in the numerator must put out this x-π in the denominator.

Hence, k must be  π^3.  Then the numerator, as x approaches to π, is equivalent 
to  π^3-x^3 = (π-x)*(π^2 +πx + x^2), and it puts out x-π in the denominator.


So, we deduced logically that the expression is a continuous function if and only if  k = π^3.


Then  kb = ,

f(π) = -  = - ;


and  kb - f(π) =  + .



It gives us this equation

     +  = 4*pi^(5/2).


and we want to solve it for "b".  Divide all the terms by  to get

     +  = .


This equation is not algebraic and can not be solved using algebraic methods.


So, solve it numerically/graphically.


I used online graphical solver www.desmos.com/calculator  
(free of charge online solver for common use).


See the plot and the numbers in it under this link  

    https://www.desmos.com/calculator/20kerz6s92

    https://www.desmos.com/calculator/20kerz6s92


ANSWER.  As you can see from the plot, there are many different possible solutions for "b".

         They are x-coordinates of intersections of the curves with the horizontal straight line. 

         There are positive and negative solutions. Of them, the minimal positive value of  "b"  is  0.62526  (approximate).

You apparently meant this, for it is another L'Hopital's theorem problem from
the limit section of Calc II.

If  is continuous at , and  , find the value of .


Since numerator and denominator both approach zero as x approaches π,
L'Hopital's theorem applies and










 









That is only valid for  and 





 and  



Answer , 

Edwin

Here are a couple of videos on L'Hopital's theorem for you:

The undefined indeterminate forms are 


https://www.youtube.com/watch?v=Gh48aOvWcxw

https://www.youtube.com/watch?v=g45quDrSA6g


Edwin

Ikleyn took the denominator as

(sin(b)×(x-π)),  as it was typed.  

However, I purposely changed it to 

(sin(b(x-π)).

She tried to solve this problem using f(x) as it was written:



and found that it was unsolvable for b by ordinary mathematical means.

However, I assumed the student had mistyped, and that f(x) should have been
this instead: 



By making this change, I was able to solve for b using the quadratic formula.

Why did I assume that the student has mistyped?

I reasoned that no professional math problem-creator would have included 
only one sine function of a constant, in this case, sin(b).  If a professional
problem-creator included a sine function, they certainly would have used a sine
function of a VARIABLE, not the sine of a CONSTANT, which would be a constant
itself!  So I changed the denominator so that the sine function would be of a
variable, not of a constant.

Also, no professional math problem-creator would create a problem that would
not be solvable by ordinary means taught in normal calculus courses.

So I changed the denominator (sin(b)×(x-π)) to (sin(b(x-π)) and was able to
solve for b using the quadratic formula.

It made more sense to me that a typo had been made, rather than an incompetent
math-problem-creator -- especially since there was a typo of a missing right
parenthesis in this: 

4(π^(5/ 2    

which Ikleyn mentioned.

Also, when a problem contains k, this letter is nearly always taken as an
arbitrary constant, but Ikleyn found k not to be arbitrary at all.  

I think my interpretation was the intended problem.  Maybe Ikleyn will agree
with me on that.  

Edwin