Question (1 Viewer)

[tacogym27101990]

When you differentiate an equation you are pretty much finding the tangent i believe

finding the gradient of the tangent

 

[shaon0]

Thanks for the links.
I'll post my answers for the first part.

1.
a) 2x
b)3
c)-2x
d)4
e)-1

 

[shaon0]

I just realised, Can't you just get the index of each pronumeral and multiply them by the co-effecients respectively.
Then you can decrease each pronumerals index by one.

 

[Iruka]

Yep, that's how you differentiate a polynomial.

It's beautiful, really. It all comes out of Pascal's triangle.

 

[shaon0]

Yep, that's how you differentiate a polynomial.

It's beautiful, really. It all comes out of Pascal's triangle.

Yea your right. If i read a little faster i would have found out

 

[shaon0]

How can i find the f'(3)?

 

[Continuum]

I just realised, Can't you just get the index of each pronumeral and multiply them by the co-effecients respectively.
Then you can decrease each pronumerals index by one.

Yeah, that's the easier way of differentiating stuff.

 

[conics2008]

How can i find the f'(3)?

to find F prime (3)

first you need to find dy/dx and then sub the value 3 into the function.

 

[shaon0]

awesome! you just differentiated from first principles



eg. for a) you have f ' (x) = 2x, to get f ' (3) you sub in 3 for x.

a) f ' (3) = 2(3) = 6
b) f ' (3) = 3
c) f ' (3) = - 2(3) = -6
etc

given some equation say, f(x) = x^2 (parabola), f ' (3) will give the gradient of the tangent at the point where x = 3.
f (x) = x^2
f ' (x) = 2x
f ' (3) = 6
.'. the gradient of the tangent which touches the point (3,9) on the parabola has a gradient of 6

Ok. Thanks for the help. Thats easy.
Differentiaition is not that hard

 

[shaon0]

Is integration much harder?
It looks much harder than differentation.
2
a)6(3)^2 -4
=196-4
=192
b)14(x)
=14(-4)
=-56
c)-6x
=24
d)6
e)12x^2 - 2x +4
=12 (-5)^2 -2(-5)+4
=300 +10+14
=314

 

[tacogym27101990]

nah its not too hard really
basic integration is just goin the other way, really
it can get a bit complicated in 4 unit tho,

 

[shaon0]

nah its not too hard really
basic integration is just goin the other way, really
it can get a bit complicated in 4 unit tho,

It looks pretty complicated in 3unit.

 

[shaon0]

integration is slightly harder, but its very similar - its the reverse process of differentiation!

say we have f(x) = x

to find the integral F(x), we reverse the process, i.e. plus 1 to the power (instead of minus) and divide by the new power (instead of multiply)

so, F(x) = x^2/2 + C (C is just a constant you add after you integrate)

now when we differentiate this we get back to x easy

awesome stuff by the way! good to see youre having a go by yourself at a new topic

Thanks

 

[shaon0]

Answers for Part 3.
1
a) x=7/2
b)-2
2
a) -1
b)16
c)2

 

[tacogym27101990]

It looks pretty complicated in 3unit.

thats only cause havn't learnt it yet
see how quickly u picked up differentiation
ull be fine with it

 

[shaon0]

YES! Now i know how to differentiate.

 

[shaon0]

1. a)
f(x) = -2x^2 +6x - 4
f ' (x) = -4x + 6
20 = -4x + 6
14 = - 4x
x = - 7/2
just dropped the negative

question 2 is a bit different, theyre after stationary points. stationary points are where the derivative is equal to 0 (f ' (x) = 0). for now dont confuse yourself with the different types of stationary points, just know what it means: when we find a stationary point, we're finding a point on the graph where the gradient of the tangent is equal to 0.

eg.
2. a)
f(x) = x^2 + 6x + 8
f ' (x) = 2x + 6
f ' (x) = 0 to find where the gradient of the tangent is equal to 0
0 = 2x + 6
x = -3

this means on the graph of y = x^2 + 6x + 8 (which is a parabola), at the point where x = -3, the gradient of the tangent to the graph is 0.
in other words, at the point (-3, -1) on the parabola we have a horizontal tangent.
if you think about the significance of this point, you'll realise its the "hump" of the parabola: View attachment 16751

Ok i'll try again.
Thanks

 

[Continuum]

1. a)
f(x) = -2x^2 +6x - 4
f ' (x) = -4x + 6
20 = -4x + 6
14 = - 4x
x = - 7/2
just dropped the negative

question 2 is a bit different, theyre after stationary points. stationary points are where the derivative is equal to 0 (f ' (x) = 0). for now dont confuse yourself with the different types of stationary points, just know what it means: when we find a stationary point, we're finding a point on the graph where the gradient of the tangent is equal to 0.

eg.
2. a)
f(x) = x^2 + 6x + 8
f ' (x) = 2x + 6
f ' (x) = 0 to find where the gradient of the tangent is equal to 0
0 = 2x + 6
x = -3

this means on the graph of y = x^2 + 6x + 8 (which is a parabola), at the point where x = -3, the gradient of the tangent to the graph is 0.
in other words, at the point (-3, -1) on the parabola we have a horizontal tangent.
if you think about the significance of this point, you'll realise its the "hump" of the parabola: View attachment 16751

:rofl: