Question

Sketch the given curve, using the following steps.

Y(x) = 2√x - x
A) Domain
B) Intercepts
C) Asymptotes
D) Intervals of increase and decrease + local min and max.
E)Concavity + Inflection point.

Answer 1

This is what I got:
A) Domain: (-∞,0) U (0,∞)
B) Y-Int = 0
x-Int = 0
C) Increasing at (1,∞)
Decreasing at (-∞,1)
Local Min: f(1) = 1
No Local Max
f) No concavity + inflection point.

Answer 2

Please help!

Answer 3

is this the y? : \(\large y(x)=2\sqrt{x}-x \)

Answer 4

Yup

Answer 5

I'm not sure about the local min/max and the concavity bit. Also, when I graphed this in my calculator, it showed a point of intersection at (3.8, 0). So I'm really confused.

Answer 6

Oh, I forgot to include asymptotes to the list. I got no asymptotes.

Answer 7

the domain is [0, \(\large \infty \))

Answer 8

Okay, what about (-∞,0)?

Answer 9

for the x intercepts, set y=0:
\(\large 0=2\sqrt{x}-x\rightarrow x^{\frac{1}{2}}(2-x^{\frac{1}{2}})=0 \)
so x=0, x=sqrt2 are your x intercepts.

Answer 10

y-intercept is at y=0

Answer 11

Okay.

Answer 12

you can't stick in any negative values of x into that function... so the domain is [0, infinity)

Answer 13

Oh, because of the square root? Okay, got it.

Answer 14

yep...

Answer 15

no asymptotes....

Answer 16

Okay, that's what I got too.

Answer 17

\(\large y(x)=2\sqrt{x}-x \)
\(\large y(x)=2x^{\frac{1}{2}}-x \)
\(\large y '(x)=2(\frac{1}{2})x^{-\frac{1}{2}}-1\rightarrow 0=2(\frac{1}{2})x^{-\frac{1}{2}}-1 \)
\(\large x^{-\frac{1}{2}}-1=0 \)
\(\huge \frac{1}{x^{\frac{1}{2}}}=1 \rightarrow x=1\)

Answer 18

so test the function for increasing/decreasing using the first derivative test on the intervals (0, 1) , (1, infinity)

Answer 19

Okay, so I got increasing at (1,∞) and decreasing at (0,1)?

Answer 20

Is that okay?

Answer 21

you got that backwards.... it should be increasing (0, 1), decreasing (1, infinity)

Answer 22

Oh, my bad!

Answer 23

So, there's a local max at at x=1 then?

Answer 24

zero is part of the domain.. so x=0 is a local min.
local max at x=1

Answer 25

Okay. Also, just wanted to double check, you plug in the values into the original equation to find the local min/max, right?

Answer 26

you can use the first derivative test.... to the left of x=1, y is increasing, to the right of x=1, y is decreasing... so at x=1 is a local max

Answer 27

Oh okay. I get it.

Answer 28

same with x=0... but you can only test on the right side of x=0....

Answer 29

Gotchya! :)

Answer 30

So, this is what I got as my second derivative f"(x) = -1/2X^(-3/2). When I equate f"(x) = 0, I get 0, concave downward at x=0, no point of inflection.