Basic calculus question, Leaving Cert

eshortie

This is my first time visiting the maths forum here on boards, thank ggod there is one, I'm really stuck here revising.

Can anyone help me with part two of the question below, it would be easy with a more simple equation without a power, I just can't figure ou this one due to the power of 4. All help appreciated


f(x) = (5x-2)to power of 4

(i) Find f'(x) the derivative of f(x)........easy

(ii) Find the co-ordinates of the point on the curve y=f(x) at which the slope of the tangent is 20......not so easy


Thanks, any help appreciated

LeixlipRed

If you differentiate the curve y=f(x) you get an expression for the derivative dy/dx. dy/dx is the slope of the tangent to the curve at some point x. So if you know the slope of a particular tangent is 20 and you want to know what point it occurs at let dy/dx = 20 and solve for x.

eshortie

Thanks LeixLipRed I did that, but the differentiated version is
dy/dx = 20(5x-2) to the power of 3........so

20(5x-2) to power of 3 = 20

I get lost when I have to do (5x-2)(5x-2)(5x-2) = 20

I keep getting the wrong answer I think.

Johnny86

eshortie wrote: »

This is my first time visiting the maths forum here on boards, thank ggod there is one, I'm really stuck here revising.

Can anyone help me with part two of the question below, it would be easy with a more simple equation without a power, I just can't figure ou this one due to the power of 4. All help appreciated


f(x) = (5x-2)to power of 4

(i) Find f'(x) the derivative of f(x)........easy

(ii) Find the co-ordinates of the point on the curve y=f(x) at which the slope of the tangent is 20......not so easy


Thanks, any help appreciated

so for part (ii)

y=(5x-2)^4
start with x=0, x=1 then x=2 etc, each time giving u a value for y

and you will then get many coordinates for x and y for which you can plot this curve. prob just need the first couple of points.
then find on the graph where y=20, get the x co-ord and thats the coordinates..the x cord is prob 1.2 or 1.3 i guess.

at least i think this is correct..im open to anyone who disagrees

Johnny86

actually i think im wrong..i just got you where y=20..not the slope of the tangent =20

eshortie

Yeah as LeixlipRed said first get dy/dx then put it equal to 20.

so (5x-2)to power of 3 = 20

Stuck there ha.

Answer should be (3/5 , 1) according to the back of my papers.

LeixlipRed

Yeh, that's completely wrong Johnny.

So, dy/dx = 20(5x-2)^3.

Let dy/dx = 20

so you have,

20(5x-2)^3 = 20

or

(5x-2)^3 = 1

Not 20 as you said.

eshortie

eshortie wrote: »

This is my first time visiting the maths forum here on boards, thank ggod there is one, I'm really stuck here revising.

Can anyone help me with part two of the question below, it would be easy with a more simple equation without a power, I just can't figure ou this one due to the power of 4. All help appreciated


f(x) = (5x-2)to power of 4

(i) Find f'(x) the derivative of f(x)........easy

(ii) Find the co-ordinates of the point on the curve y=f(x) at which the slope of the tangent is 20......not so easy


Thanks, any help appreciated

You'll get 20(5x-2)^3=20,
Divide both sides by 20: (5x-2)^3 = 1
Get the cubic root of both sides: (5x-2) = 1
Then just solve algabraically, 5x - 2 = 1; 5x = 3; x = 3/5

So the point is:

y = (5x-2)^4; y = 1^4; (3/5, 1)

Edit: too late.

Edit 2: Oops, just noticed you updated the Charter a few weeks ago, LeixlipRed. Haven't been around here in a few months, so I apologise for giving out a full solution.

eshortie

Same mistake as OP?

Yeah thanks everyone I just didn't realise about the cubic root, I figured the 20's would cancel. Thanks so much, relaxed a bit now again. Haha

Thanks again everyone really just want a B3 in ordinary

Iwasfrozen

f(x) = (5x-2)to power of 4

f(x)=(5x-2)^4
f'(x)=4(5x-2)^3(5)=4[(25x^2+20x+4)(5x-2)]5=20[125x^3-50x^2+100x^2-40x+20x-8]=20[125x^3+50x^2-20x-8].
When tangent is equal to 20.
20[125x^3+50x^2-20x-8]=20
125x^3+50x^2-20x-8=1
125x^3+50x^2-20x-9=0

Iwasfrozen

Iwasfrozen wrote: »

f(x)=(5x-2)^4
f'(x)=4(5x-2)^3(5)=4[(25x^2+20x+4)(5x-2)]5=20[125x^3-50x^2+100x^2-40x+20x-8]=20[125x^3+50x^2-20x-8].
When tangent is equal to 20.
20[125x^3+50x^2-20x-8]=20
125x^3+50x^2-20x-8=1
125x^3+50x^2-20x-9=0

There isn't any need to actually work out the power. See my post.

Iwasfrozen

oh yeah, I see.