Limit Problem: f(x) decreases without bound as x approaches what value from the right?

Question

This is for a calculus class I am currently taking but I do not have this math class until spring break is over and I am unsure of what this question is asking for.


f(x) decreases without bound as x approaches what value from the right?

f(x) = 6 / ( x - 2 ) ( 7 - x )

x -> ?


Any explanation is welcomed instead of the answer itself. Thank you!

Answer 1

I assume that (x-2) and (7-x) are both on the denominator here.

Obviously the critical points are at x = 2 and x = 7.

When x is just above 2, (x-2) is positive and (7-x) is positive. So as x approaches 2 from the right, the denominator is approaching 0 from above and the result is increasing without bound (since the numerator is positive).

When x is just above 7, (x-2) is positive and (7-x) is negative. So as x approaches 7 from the right, the denominator is approaching 0 from below and the result is decreasing without bound (since the numerator is positive).

So the answer is 7. (Note that the function also decreases without bound as x approaches 2 from the left.)

Answer 2

"Decreases without bound" means we're looking for a place on the graph where f(x) drops to negative infinity. In other words, a vertical asymptote.

Potential discontinuities in the function occur where the function is undefined, at values of x that are not in the domain.

Since we can't divide by zero, that will be when x-2=0 or when 7-x=0.
Solving each, we find that x=2 and x=7 are not in the domain of f(x).

The question now is, which of these decreases without bound as x approaches from the right. I imagine there is a more rigorous way to do it, but I'm lazy and would just plug in numbers close to but slightly greater than 2 and 7 and see which ones become increasingly large and negative.

Good luck!

Answer 3

The answer is 7 because as x approaches 7 from the right, the product (x-2)(7-x) is negative and becomes more and more so as x gets closer and closer to 7.