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

In summary, the author got a question on a Curve sketching assignment and is confused about some of the concepts. F(x) = x2-4x1/2 is the function being evaluated, and x is the input. There is no symmetry about the X or Y axes, and the function increases when F'(x) >0 and decreases when F'(x) <0. The max and min values can be found by finding the critical points, which are points at which the concavity changes.
  • #1
Mspike6
63
0
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 [tex]\sqrt{x}[/tex]

a) State the domain
It's domain is set for all positive real numbers


b) Determine the intercepts.
X-Intercept.

x2-4 [tex]\sqrt{x}[/tex]=0

x4 - 16x=0
x=0 or x= [tex]\sqrt[3]{16}[/tex]


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) =x2-4x1/2

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

The Function increases when F'(x) >0

2x-2x-1/2>0

2x-1/2(x3/2-1) >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(x3/2-1) <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
 
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  • #2
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.
 
  • #3
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 [itex]x^2- 4\sqrt{x}= 0[/itex] and then, immediately, [itex]x^4- 16x= 0[/itex]. The intermediate steps should be [itex]x^2= 4\sqrt{x}[/itex] and [itex]x^4= 16x[/itex].
 
  • #4
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
 
  • #5
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.
 
  • #6
jegues said:
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)
 
  • #7
Mspike6 said:
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)
 
Last edited by a moderator:
  • #8
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 .
 
  • #9
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.
 
  • #10
Mspike6 said:
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)
[tex]f(x)= x^2- 4\sqrt{x}= x^2- 4x^{1/2}[/tex]
[tex]f'(x)= 2x- 2x^{-1/2}[/tex]
[tex]f"(x)= 2+ x^{-3/2}[/tex]

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 .
 

1. What is the function F(x) = x2-4 \sqrt{x}?

The function F(x) = x2-4 \sqrt{x} represents a mathematical relationship between an input value x and an output value y. It is a combination of two basic functions, x2 and 4 \sqrt{x}, and the result is a curve that can be graphed.

2. How do I sketch the curve of F(x) = x2-4 \sqrt{x}?

To sketch the curve of F(x) = x2-4 \sqrt{x}, you can use a graphing calculator or plot points on a graph. Choose a range of values for x, plug them into the function, and plot the resulting points. Then, connect the points to create a smooth curve.

3. What is the domain of F(x) = x2-4 \sqrt{x}?

The domain of a function refers to all the possible input values. In the case of F(x) = x2-4 \sqrt{x}, the domain would be all real numbers greater than or equal to 0, since the square root function cannot take negative inputs.

4. What is the range of F(x) = x2-4 \sqrt{x}?

The range of a function refers to all the possible output values. In the case of F(x) = x2-4 \sqrt{x}, the range would be all real numbers less than or equal to 0, since the function will never output a positive value due to the square root function.

5. Can I use calculus to analyze the function F(x) = x2-4 \sqrt{x}?

Yes, you can use calculus to analyze the function F(x) = x2-4 \sqrt{x}. You can find the derivative of the function to determine the slope of the curve at any point and the second derivative to determine the concavity of the curve. This can provide valuable information about the behavior of the function.

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