Hey mathletes I need help with question 17. Anyone know how to

it's indeterminate at the value x=2, which means it's 0/0. Take the derivative of the top and the derivative of the bottom evaluated at x=2

i tought my math was difficult, damn

>>723405199
nice b8 m8

you should really have no trouble with 17 if you were able to do any of the other problems, all of these limits involve indeterminate functions

>>723404904
the function f has two possible outputs. an equation or k. Therefore the function is continuous if k equals the equation.

Probably you should put -(2/7) as x into the equation an chose the result as k.

>>723405703
Not necessarily. Some of them you can just factor out some (x-a) to get rid of the singularity in the denominator

>>723405864
okay. doesn't work. You can't do it.

gr8 b8 m8

>>723404904
Pls add sets to the top of the function

Answer is 8. Multiply top and bottom by the conjugate of the denominator and it's easy after that

Thank everyone for replying

>>723404904
>Answer is 8. Multiply top and bottom by the conjugate of the denominator and it's easy after that
Thank you :)

As said, in x=2, the fraction is undefined (otherwise, it would give 0/0). So you want to know what the fraction tends to when x approaches 2. Then you would make k equal this limit.

The way to do it here is to multiply both the numerator and denominator by:

√(7x+2) + √(6x+4)

The denominator becomes:

(7x+2) - (6x+4) = x-2

which cancels out with the x-2 of the numerator, leaving:

f(x) = √(7x+2) + √(6x+4), if x≥-2/7 and x≠2

Calculate the limit of this function when x approaches 2: you get 4+4 = 8, your value of k.

>>723408566

Fuck! Took me too long to find those stupid symbols and write the whole thing. Hope it helps, though.