Sketching the Curve of the Function F(x): x2-4 \sqrt{x}

[Mspike6]

Hello,
I got this lobe Curve sketching question on my assignment. i know most of it but i think i got the drevetive part wrong. anyway, i will write the whol question with all the soultions i reached.

1] Use the Funtion F(x) = x²-4 \sqrt{x}

a) State the domain

It's domain is set for all positive real numbers

b) Determine the intercepts.

X-Intercept.

x²-4 \sqrt{x}=0

x⁴ - 16x=0
x=0 or x= \sqrt[3]{16}


Y-intercept

y= (0)²-4(0)
Y intercept is 0

c)Is the graph Symmetric about the X-axis or Y axis ?

No, There is no symmetry since the equation will change if we replaced (x,y) by (-x,y) or (x,-y)

d) Find the Asymptotes

There is no Asymptotes

e) where does the function increase ? where does the function decrease ?

F(x) =x²-4x^1/2

F'(c) = 2x-2x^-1/2

The Function increases when F'(x) >0

2x-2x^-1/2>0

2x^-1/2(x^3/2-1) >0

x > 0 or x > 1

The curve is Increaseing in (1, Invfinity)


The function Decreases when F'(x) <0
2x-2x^-1/2<0

2x^-1/2(x^3/2-1) <0

x < 0 or x < 1

The curve is Decreasing (0, 1)

f. Determine the Maximum and minimum values.

From here it all start making no sens. that's why i think i made a mistake in the drevative process.

But to get the maximum and minimum, i think i should get the Critical Points[ F'(x) =0 ] and then test those points on the second drevative test,
If F''(x) > 0 Then x is a minimum
If F'' (x) < 0 then x is a maximum
If F'' (x) = 0 then the test failed.


I got F''(X) to be equal to 2+x^-3/2

Please, if someone could tell me which part is wrong, and direct me in the right way!

Thank, you

 

[Vaal]

With the exception of a one small issue in part e ("x > 0 or x > 1" should just be x > 1) you have everything correct including your derivatives.

You can find the max and min values with the critical points you have from part e, x=0 and x=1, in the manner that you specified above. Just a heads up, x=0 is a little trickier because it's the end of the domain but if you sketch a rough graph you should be able to figure it out.

 

[HallsofIvy]

The domain is "all non-negative real numbers" which is slightly different from "all positive real numbers".

Also while your values for the x-intercepts are correct, your method is peculiar. You have x^2- 4\sqrt{x}= 0 and then, immediately, x^4- 16x= 0. The intermediate steps should be x^2= 4\sqrt{x} and x^4= 16x.

 

[Mspike6]

Thank you so much Vaal and HallsOfIvy , really appreciate you insights


I drew The graph of the function on my graphic calcu.

and both 0 has y-value of 0 and 1 has y-value of -3 .
but the Second derivative test shows that bpth are minimums (F"(x) > 0) , which makes no sense.


F"(X) = 2+x^-3/2


F"(0) = 2 (since is't greater then 0 it's a minimum)
F"(1) = 3 (Since it's greater then 0 it's a minimum )

It makes sense that 1 is a minimum, but why 0 ? if someone could explain that point to me, would be really appreciated

 

[jegues]

Minimums and maximums are determined by the points of inflection.

Points of inflection are points in which the concavity changes. Assuming you've written down the second derivative correctly, when I graph it it seems to me that it is always concave up so there are no points of inflection.

 

[Mspike6]

Minimums and maximums are determined by the points of inflection.

Points of inflection are points in which the concavity changes. Assuming you've written down the second derivative correctly, when I graph it it seems to me that it is always concave up so there are no points of inflection.

Now this is getting me confused :D.

ya Inflection point is a way to know the minimums, and maximums. but the main test is to find the critical points, then do the second test.

and you the graph is always concave up except between interval(0,1)

 

[HallsofIvy]

Now this is getting me confused :D.

ya Inflection point is a way to know the minimums, and maximums. but the main test is to find the critical points, then do the second test.

No, inflection points have little or nothing to do with "maximum" and "minimum". Inflection points are points are points where the first derivative changes sign (and so the second derivative is 0). Functions that have NO minimum or maximum may have inflection points and, conversely function that do have inflection points may have no maximum or minimum.

and you the graph is always concave up except between interval(0,1)

 

[Mspike6]

Thanks HallsOfIvy , again, for your help.
my last questions is, and am sorry if i asked alot, Is 0 a Minimum ? if not , then why it's F"(0) is Greater then 0 (which means that it's minimum)

I think that 0 is a minimum, because The Second drevative test said so,

i don't think that it's a minimu, because it is the end point, and f(-0.0001) doesn;t exist .

 

[Vaal]

That is a good question. The second derivative test actually will not work on critical values that lie on the very edge of the original function's domain(i.e. the point where x=0 in this case). From the graph of the original function you can see x=0 is the highest point point in the neighborhood of x=0 but less than the value of the function as it increases towards infinity which would mean it is a relative maximum if anything. Sense x=0 is the edge of the domain it's neighborhood is not really defined (because any values to the left of x=0 aren't in the domain) so I think it actually isn't even a local minimum. Perhaps HallsofIvy can confirm this.

 

[HallsofIvy]

Thanks HallsOfIvy , again, for your help.
my last questions is, and am sorry if i asked alot, Is 0 a Minimum ? if not , then why it's F"(0) is Greater then 0 (which means that it's minimum)

f(x)= x^2- 4\sqrt{x}= x^2- 4x^{1/2}
f&#039;(x)= 2x- 2x^{-1/2}
f&quot;(x)= 2+ x^{-3/2}

No, the second derivative is NOT greater than 0 at x= 0, it is not even defined there.

For x very close to 0, f'(x) is a large negative number so f is decreasing away from x= 0 and x= 0 is a local maximum.

I think that 0 is a minimum, because The Second drevative test said so,

i don't think that it's a minimu, because it is the end point, and f(-0.0001) doesn;t exist .