Calculus Question: Find all critical points of f(X) = (2X^2 + 18 + X ) / X?

Question

f(X) = (2X^2 + 18 + X ) / X

the critical point is the derivative at 0, which is (2X^ 2 - 18 ) / X^2 = 0.

multiply both side by X^2 and you have (2X^2 - 18) = 0
divide both side by 2 and you have (X^2 - 9 ) = 0
Factor and you have (X - 3) (X + 3) = 0
So X = 3 and X = -3

According to my text book, it says that -3 is not in the domain. But WHY?? using -3 for X in the original formula does not cause any undefined values. Why is -3 not in the domain? Please let me know.

thank you for your relpy, but nyphdinmd, my derivative is not wrong. It was just initially written without simplifying. If you read further down, the simplification when setting the derivative to 0 is same as yours 0 = X^2 - 9 .

And my guess is x= -3 is in the domain as y is -11 at that point. But not according tot he textbook. You think the textbook is wrong?

Answer 1

YOu have the derivative wrong:

It is (4x+1)/x - (2x^2+18+x)/x^2 setting that to zero gives:

0 = (4x+1)/x - (2x^2+18+x)/x^2 =(4x+1) - (2x^2+18+x)/x

0 = 4x^2 +x - 2x^2-18-x =2x^2 -18 = x^2 - 9 --> x = +/- 3

The derivative evaluated at x = 0 is undefined there is a term going to infinty as 1/x^2 as x goes to zero. f(x) is discontinuous here - it blows up as x approaches zero as well.

You can ask what type of points x = +/-3 are. Tale the seconde derivative (f"(x)) and substitute for x the values +3 and -3. If f"(x) = 0 , it is an inflection point, if f"(x) < 0 it is a local maximum, and if f"(x) > 0 it is a local minimum.

f(-3) is finite - it is -11 so clearly in the range and domain of f(x).

Answer 2

f(x)=(2x^2+18+x) / x
rewrite
f(x)=2x + 18/x +1
f'(x) = 2-18/x^2 =0
2x^2-18=0
2x^2=18
x^2=9
x=+-3 (critical points)
I don't see why either!