How to differientiate these Polynomial functions

**ghostanime2001** · June 24th 2008, 06:50 PM

Hi I've been stuck with this question for quite a while. School's been over for a week and I have been trying to solve this problem over and over and over again and I failed to get the answers to these following questions. If anybody who can help me please do help I'm sure somebody out there will benefit from my response as well.

1. Let P(a,b) be a point on the curve √x + √y = 1. Show that the slope of the tangent at P is

-√b
------
√a

2. For the power functions f(x) = x^n, find the x-intercept of the tangent to its graph at point (1,1). What happens to the x-intercept as n increases without bound (n --> +OO (positive infinity) ) ? Explain the result geometrically

3. For each function, sketch the graph of y=f(x) and find an expresssion for f ' (x). Indicate any points at which the f ' (x) does not exist.

a) f(x) = [ x^2, x < 3
[ x + 6, x >-(x is greater or equal then 3) 3 ]

b) f(x) = I3x^2 - 6I I means absolute value bars

c) f(x) = I IxI - 1 I I means absolute value bars

That is all. I'd greatly appreciate it. Thanks.

**mr fantastic** · June 24th 2008, 07:50 PM

Originally Posted by ghostanime2001

Hi I've been stuck with this question for quite a while. School's been over for a week and I have been trying to solve this problem over and over and over again and I failed to get the answers to these following questions. If anybody who can help me please do help I'm sure somebody out there will benefit from my response as well.

1. Let P(a,b) be a point on the curve √x + √y = 1. Show that the slope of the tangent at P is

-√b
------
√a

Mr F says: Are you familiar with implicit differentiation?

2. For the power functions f(x) = x^n, find the x-intercept of the tangent to its graph at point (1,1). What happens to the x-intercept as n increases without bound (n --> +OO (positive infinity) ) ? Explain the result geometrically

Mr F says: Where are you stuck? ${\color{red} f'(x) = n x^{n-1} \Rightarrow m = n \, \text{at} \, x = 1}$ . Then the equation of the tangent is ${\color{red} y - 1 = n(x - 1) \Rightarrow y = ......}$ .

3. For each function, sketch the graph of y=f(x) and find an expresssion for f ' (x). Indicate any points at which the f ' (x) does not exist.

a) f(x) = [ x^2, x < 3
[ x + 6, x >-(x is greater or equal then 3) 3 ]

b) f(x) = I3x^2 - 6I I means absolute value bars

c) f(x) = I IxI - 1 I I means absolute value bars

That is all. I'd greatly appreciate it. Thanks.

3. Have you drawn the graphs? f'(x) does not exist at 'jumps' and at 'pointy bits' (for the highbrow, jump discontinuities and salient points).

**ghostanime2001** · June 24th 2008, 10:01 PM

yes i am familiar with implicit differentiation but i wanna know if there is another different method regardless if its easier or harder i want know and btw this is my intention was of solving this problem

x^1/2 + y^1/2 = 1
y^1/2 = 1 - x^1/2
y = (1 - x^1/2)^2
y = (1 - x^1/2)(1 - x^1/2)
y = 1 + 2x^1/2 + x

y' = 2(1/2)x^-1/2 + 1
y' = x^-1/2 + 1

But then here is when i get stuck i got the slope y' = x^-1/2 + 1 but how do i show it being equal to

-√b
-----
√a

Thanks again

**mr fantastic** · June 24th 2008, 10:15 PM

Originally Posted by ghostanime2001

yes i am familiar with implicit differentiation but i wanna know if there is another different method regardless if its easier or harder i want know and btw this is my intention was of solving this problem

x^1/2 + y^1/2 = 1
y^1/2 = 1 - x^1/2
y = (1 - x^1/2)^2
y = (1 - x^1/2)(1 - x^1/2)
y = 1 + 2x^1/2 + x

y' = 2(1/2)x^-1/2 + 1
y' = x^-1/2 + 1

But then here is when i get stuck i got the slope y' = x^-1/2 + 1 but how do i show it being equal to

-√b
-----
√a

Thanks again

Note that $\sqrt{b} = 1 - \sqrt{a}$ . Therefore $- \frac{\sqrt{b}}{\sqrt{a}} = ~ ...... ~ = 1 - \frac{1}{\sqrt{a}}$ ........ Capisce?

**ghostanime2001** · June 25th 2008, 10:55 AM

uhh.... im still not understanding this because we got this homework before we got into implicit differentation so thats why im asking how show it being equal to that without using implicit differention and why when i put the equation into implicit form then try to differentiate it I cannot show it being equal to that - sqrt(b)/sqrt(a)

**mr fantastic** · June 25th 2008, 11:45 AM

Originally Posted by ghostanime2001

uhh.... im still not understanding this because we got this homework before we got into implicit differentation so thats why im asking how show it being equal to that without using implicit differention and why when i put the equation into implicit form then try to differentiate it I cannot show it being equal to that - sqrt(b)/sqrt(a)

I have already shown you how!

Working backwards from the result given in the question:

Note that

. Therefore

From your own work:

x^1/2 + y^1/2 = 1
y^1/2 = 1 - x^1/2
y = (1 - x^1/2)^2
y = (1 - x^1/2)(1 - x^1/2)
y = 1 - 2x^1/2 + x

Note the mistake you made in the expansion.

y' = - 2(1/2)x^-1/2 + 1
y' = - x^-1/2 + 1

Therefore at x = a: y' = - a^-1/2 + 1

Clearly the required result is now shown.

**ghostanime2001** · June 25th 2008, 01:59 PM

I still dont see where you are going with this
Why is y' = - a^-1/2 + 1 = -sqrt(b)/sqrt(a)

Just HOW ? are there little methods to show the left side is equal to the right side ??? if i want to show the two are equal. Just bluntly stating the two are equal doesnt prove anything. I want to actually see they are equal not just putting the equal sign in between them and concluding they are equal. No.

**mr fantastic** · June 25th 2008, 05:14 PM

Originally Posted by ghostanime2001

I still dont see where you are going with this
Why is y' = - a^-1/2 + 1 = -sqrt(b)/sqrt(a)

Just HOW ? are there little methods to show the left side is equal to the right side ??? if i want to show the two are equal. Just bluntly stating the two are equal doesnt prove anything. I want to actually see they are equal not just putting the equal sign in between them and concluding they are equal. No.

You are told that to show that $y' = -\frac{\sqrt{b}}{\sqrt{a}}$ .

This expression is the same as $- \frac{(1 - \sqrt{a})}{\sqrt{a}} = - \frac{1}{\sqrt{a}} + 1$ .

This is because you are told that $\sqrt{x} + \sqrt{y} = 1$ and you are given the point (a, b). Therefore $\sqrt{a} + \sqrt{b} = 1 \Rightarrow \sqrt{b} = 1 - \sqrt{a}$ .

You have done a calculation and found (after my corrections) that $y' = - \frac{1}{\sqrt{x}} + 1$ . When you substitute x = a into this you get $- \frac{1}{\sqrt{a}} + 1$ .

I have not just bluntly stated that the two are equal. They are equal via a clear and logical process.

If you still don't get it, then I'm sorry but I'm done here ....... maybe someone else has the time and inclination.

**Mathstud28** · June 25th 2008, 05:17 PM

Originally Posted by ghostanime2001

Hi I've been stuck with this question for quite a while. School's been over for a week and I have been trying to solve this problem over and over and over again and I failed to get the answers to these following questions. If anybody who can help me please do help I'm sure somebody out there will benefit from my response as well.

1. Let P(a,b) be a point on the curve √x + √y = 1. Show that the slope of the tangent at P is

-√b
------
√a

2. For the power functions f(x) = x^n, find the x-intercept of the tangent to its graph at point (1,1). What happens to the x-intercept as n increases without bound (n --> +OO (positive infinity) ) ? Explain the result geometrically

3. For each function, sketch the graph of y=f(x) and find an expresssion for f ' (x). Indicate any points at which the f ' (x) does not exist.

a) f(x) = [ x^2, x < 3
[ x + 6, x >-(x is greater or equal then 3) 3 ]

b) f(x) = I3x^2 - 6I I means absolute value bars

c) f(x) = I IxI - 1 I I means absolute value bars

That is all. I'd greatly appreciate it. Thanks.

For the first one we have two choices

$\sqrt{x}+\sqrt{y}=1\Rightarrow{y=(1-\sqrt{x})^2}$

Now just differentiate, or you can do this

$\frac{1}{2\sqrt{x}}+\frac{y'}{2\sqrt{y}}=0$

Now plug in x and y and solve for y'

**ghostanime2001** · June 26th 2008, 08:14 PM

If i differentiate using your first way Mathstud28
y = (1- √x)²
= (1- √x)(1- √x)
= 1 - 2√x + x

f'(x) = -2(1/2)x^-½ + 1
= -x^-½ + 1
= -1/√x + 1

substituting x->a and y'->? i get this

? = -1/√a + 1

How do i from here show the slope of tha tangent at that √x + √y = 1 at P(a,b) is

-√b
-----
√a

The thing to remember what im sayin here is that this worksheet i got in school, was given before the class learned implicit differentiation. So That's my problem here how do i prove it equals that without having to even think about implicit differentiation, if i was a currently enrolled calc student in class that didnt know what implicit differentiation meant.

Thanks again

**mr fantastic** · June 26th 2008, 08:48 PM

Originally Posted by ghostanime2001

If i differentiate using your first way Mathstud28
y = (1- √x)²
= (1- √x)(1- √x)
= 1 - 2√x + x

f'(x) = -2(1/2)x^-½ + 1
= -x^-½ + 1
= -1/√x + 1

substituting x->a and y'->? i get this

? = -1/√a + 1

How do i from here show the slope of tha tangent at that √x + √y = 1 at P(a,b) is

-√b
-----
√a

The thing to remember what im sayin here is that this worksheet i got in school, was given before the class learned implicit differentiation. So That's my problem here how do i prove it equals that without having to even think about implicit differentiation, if i was a currently enrolled calc student in class that didnt know what implicit differentiation meant.

Thanks again

Read post #8 again.

**ghostanime2001** · June 26th 2008, 09:37 PM

So im just trying to prove Left side equals the right side then to show the slope ???

**mr fantastic** · June 26th 2008, 09:57 PM

Originally Posted by ghostanime2001

So im just trying to prove Left side equals the right side then to show the slope ???

Yes.

**Jhevon** · June 26th 2008, 10:00 PM

Originally Posted by ghostanime2001

So im just trying to prove Left side equals the right side then to show the slope ???

i am sorry. i do not understand your question, nor do i see where you're getting stuck. perhaps you can be more clear on what is confusing you. we are still talking about question 1, right?

we have shown that the slope is $1 - \frac 1{\sqrt{a}}$ , by solving for y and differentiating regularly. i think you followed that ok, right?

now, the problem asked that we must prove that the slope is $- \frac {\sqrt{b}}{\sqrt{a}}$ . thus, it remains to show that $- \frac {\sqrt{b}}{\sqrt{a}} = 1 - \frac 1{\sqrt{a}}$ .

this is what Mr F did in post #8. the complete solution is there.

we know the point (a, b) is on the curve (that is, when x = a, y = b). so, since the curve is $\sqrt{x} + \sqrt{y} = 1$ we can plug in x = a and y = b, and it will be on the curve. thus we get $\sqrt{a} + \sqrt{b} = 1$ . solving for $\sqrt{b}$ , we get, $\sqrt{b} = 1 - \sqrt{a}$

ok, now back to what we know the slope is. we know it is $1 - \frac 1{\sqrt{a}}$ . now for some algebraic manipulation

$\underbrace{1 - \frac 1{\sqrt{a}}}_{\text{our slope}} = \frac {\sqrt{a} - 1}{\sqrt{a}}$ ..............i just added the fractions

............ $= \frac {-( {\color{red} 1 - \sqrt{a}})}{\sqrt{a}}$ ...........

what's in red is exactly what $\sqrt{b}$ is! how lucky!

............ $= - \frac {{\color{red}\sqrt{b}}}{\sqrt{a}}$

as was to be shown.

**ghostanime2001** · June 26th 2008, 11:07 PM

HOly **** !?!?!?!?!?!??!?!

You almost finished my quest though T_T

But just one more question is it possible to take out the negative sign out before getting to that step so the approach to the answer is more clear and logical or is that it ??

Btw... how do u guys get that text i want to like make my own online source of solving homework questions and post it on the web so its a whole lesson from what i've learned posted on the net for the rest of the world. I'm just wondering cuz that text you guys wrote, all those mathematical symbol is so clear and nice i wanna know if you can do that in Word or something or possibly download a program that types that text. Thanks

Thread: How to differientiate these Polynomial functions

LinkBack

Thread Tools

Display

How to differientiate these Polynomial functions

Similar Math Help Forum Discussions

Polynomial Functions

Polynomial functions

Polynomial Functions help please!!!

Polynomial Functions

Odd Polynomial Functions

Search Tags