Question

The horizontal tangent lines of f(x) occur at which x-values?

Answer 1

\[f(x)=\frac{ \sqrt{2x-1} }{ x }\]

Answer 2

I got x=2/3

Answer 3

Did you take first derivative then set that equal to zero?

Answer 4

yes. I only made the numerator equal to 0

Answer 5

I got 2/3 and 0 but 0 can't be an option because it is not in the domain of f(x)

Answer 6

That all seems to check out.
I'm going to try graphing it just to be sure, but I think you did everything fine here.

Answer 7

:)

Answer 8

The graph is showing something different. You might want to double-check everything.
It looks a little funny, so might take some triple-checking too.

Answer 9

ruh roh.

Answer 10

Ah, I think I may have seen where we both made an error. Look where you need chain rule on (2x-1)^(1/2)

Answer 11

f'(x) = \[\frac{ \frac{ 1 }{ 2 } \times (2x-1)^{-2/3} \times 2 \times x-(2x-1)^{1/2}}{ x ^{2} }\]

Answer 12

right.

Answer 13

I just made that top portion equal to 0. so \[\frac{ x }{ \sqrt[3]{(2x-1)^{2}}} - \sqrt{2x-1}\]

Answer 14

I found a common denominator.

Answer 15

Why (2x-1)^(-2/3) ?

Answer 16

oh woops

Answer 17

I meant ^-1/2 :P

Answer 18

Ya. Maybe try
\[\large f(x) = (2x-1)^{1/2} \cdot x^{-1}\]
and using product rule to verify?

Answer 19

(I trip up using quotient rule too often. Product rule seems less prone to error)

Answer 20

ok. so \[\frac{ 1 }{ 2 } \times (2x-1)^{\frac{ -1 }{ 2 }} \times 2 \times x ^{-1}+ (-x ^{-2})(2x-1)^{\frac{ 1 }{ 2 }}\]

Answer 21

That should simplify a little cleaner. Yes, solve that =0.

Answer 22

\[\frac{ 1 }{ x(2x-1)^{\frac{ 1 }{ 2 }} }+\frac{ (2x-1)^{\frac{ 1 }{ 2 }} }{ x ^{2} }\]

Answer 23

ok so

Answer 24

\[\frac{ 1 }{ x(2x-1)^{\frac{ 1 }{ 2 }} }=-\frac{ (2x-1)^{\frac{ 1 }{ 2 }} }{ x ^{2} }\]

Answer 25

\[1=\frac{ 2x-1 }{ x }\]

Answer 26

\[x=2x-1\]

Answer 27

-x=-1

Answer 28

x=1?

Answer 29

That's one way of looking at it, yes. That should work out...

Answer 30

bah ok.

Answer 31

Yes, because, as you said, x=0 is not in the domain.

Answer 32

right

Answer 33

Yeah, looks like we both made the same error at first. I got 2/3 too, but then went back and saw I forgot to multiply by 2 somewhere.
*shrug* Got to be patient with this stuff. Haste makes waste and all that.

Answer 34

Well thanks!

Answer 35

arithmetic is my arch nemesis.

Answer 36

Any time.
I just told someone a couple minutes ago, "The hardest part about Calculus is the algebra."
Now I get to say, "The hardest part about algebra is the arithmetic."

:-P

Answer 37

haha. you told me that.

Answer 38

Oh yeah, on Steel11's implicit differentiation question.
X-D

Answer 39

mhmm

Answer 40

And you know what? I remind myself of this pretty often, "The hardest part about arithmetic is the counting."
Grrr, the counting! Messes me up every time!

Answer 41

Someone should come up with, I don't know, mathematics, to make all this counting easier!

.....

Oh, right.

Answer 42

psh.

Answer 43

I want to go back to the old days and use a handy dandy abacus