Critical points, increasing, and decreasing.

[Lividtea]

Consider the curve given by the function \(\displaystyle \, f(x)\, =\, \dfrac{(1\, -\, 2x)(x\, -\, 2)}{(x\, -\, 1)^2}\)
Find all critical points of f(x).

\(\displaystyle f'(x)\, =\, -\dfrac{x\, +\, 1}{(x\, -\, 1)^3}\)

I understand that critical points occur when f'(x)=0 or undefined.

derivative equal to zero:

I set f'(x) = 0, to get x=-1 (which is correct).

\(\displaystyle f'(x)\, =\, 0\, =\, -\dfrac{x\, +\, 1}{(x\, -\, 1)^3}\)

\(\displaystyle 0\, =\, -x\, -\, 1\)

\(\displaystyle x\, =\, -1\) (which is correct)

derivative is undefined:

I also have another critical point at x=1. If you let x = 1 in the function, then the denominator would be 0, and therefore the function is undefined....

\(\displaystyle (x\, -\, 1)(x\, -\, 1)(x\, -\, 1)\)

\(\displaystyle x\, -\, 1\, =\, 0\)

\(\displaystyle x\, =\, 1\) (which is wrong)

But the answer key doesnt have this as their answer.

 

[Steven G]

Consider the curve given by the function \(\displaystyle \, f(x)\, =\, \dfrac{(1\, -\, 2x)(x\, -\, 2)}{(x\, -\, 1)^2}\)
Find all critical points of f(x).

\(\displaystyle f'(x)\, =\, -\dfrac{x\, +\, 1}{(x\, -\, 1)^3}\)

I understand that critical points occur when f'(x)=0 or undefined.

derivative equal to zero:

I set f'(x) = 0, to get x=-1 (which is correct).

\(\displaystyle f'(x)\, =\, 0\, =\, -\dfrac{x\, +\, 1}{(x\, -\, 1)^3}\)

\(\displaystyle 0\, =\, -x\, -\, 1\)

\(\displaystyle x\, =\, -1\) (which is correct)

derivative is undefined:

I also have another critical point at x=1. If you let x = 1 in the function, then the denominator would be 0, and therefore the function is undefined....

\(\displaystyle (x\, -\, 1)(x\, -\, 1)(x\, -\, 1)\)

\(\displaystyle x\, -\, 1\, =\, 0\)

\(\displaystyle x\, =\, 1\) (which is wrong)

But the answer key doesnt have this as their answer.

I also have another critical point at x=1. If you let x = 1 in the function, then the denominator would be 0, and therefore the function is undefined.... Which function are you speaking about (f or f' )??
As you correctly stated the critical points are x-values which make f' (x) = 0 or undefined. So why do you say x=1 is not a cp? It is!

 

[Lividtea]

I also have another critical point at x=1. If you let x = 1 in the function, then the denominator would be 0, and therefore the function is undefined.... Which function are you speaking about (f or f' )??
As you correctly stated the critical points are x-values which make f' (x) = 0 or undefined. So why do you say x=1 is not a cp? It is!

I meant to say F'. Wolfram also agrees with the answer key that cp is at x=-1 only :/

 

[Steven G]

I meant to say F'. Wolfram also agrees with the answer key that cp is at x=-1 only :/

f ' NOT F' ! Try curve sketching the graph using 1st and 2nd derivatives without using the cp at x=1 and see how that goes. There is no question about it, x=1 is a cp.