SOLUTION: Example: Please help me solve this equation: if f(x)={{{ (3x^2+x)/2x }}} for x=!0 {{{f(0)=k}}} and if ƒ is continuous at x = 0,

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Example: Please help me solve this equation: if f(x)={{{ (3x^2+x)/2x }}} for x=!0 {{{f(0)=k}}} and if ƒ is continuous at x = 0,      Log On


   



Question 1044993: Example: Please help me solve this equation: if f(x)=+%283x%5E2%2Bx%29%2F2x+ for x=!0
f%280%29=k
and if ƒ is continuous at x = 0, then k =
A) -3/2
(B) -1
(C) 0
(D) 1
(E) 3/2 answer here

Found 2 solutions by Boreal, ikleyn:
Answer by Boreal(15235) About Me  (Show Source):
You can put this solution on YOUR website!
L'Hôpital's Rule
derivative of this is
(1/4x^2)[(2x)(6x+1)-(3x^2+x)*2}
=(1/4x^2){12x^2+2x-6x^2-2x}
=(1/4x^2)(6x^2)=6/4=3/2
E

Answer by ikleyn(52798) About Me  (Show Source):
You can put this solution on YOUR website!
.
Example: Please help me solve this equation: if f(x)=+%283x%5E2%2Bx%29%2F2x+ for x=!0
f%280%29=k
and if ƒ is continuous at x = 0, then k =
A) -3/2
(B) -1
(C) 0
(D) 1
(E) 3/2 answer here
~~~~~~~~~~~~~~~~~~~~~

The formulation is incorrect. (And the solution is incorrect as well).

The correct formulation and solution follow:


    You are given a function  f(x) = %283x%5E2%2Bx%29%2F%282x%29.
    Find the limit of the function at x-->0.


Solution. 


Factor x out parentheses in the numerator. Then cancel factor "x" in the numerator an denominator. You will get

lim   %283x%5E2%2Bx%29%2F%282x%29 =  lim  %28x%2A%283x%2B1%29%29%2F%282x%29 =  lim   %283x%2B1%29%2F2 = 1%2F2.
x-->0              x-->0               x-->0

Notice that this result  lim = 1%2F2  is not in the list.
Nevertheless, this result is correct.
Everything in the list is incorrect.

The plot illustrates it:




Plot y = %283x%5E2%2Bx%29%2F%282x%29