Differentiation (1 Viewer)

[rawker]

Q1 curve y=x^3 -2x^2 -4x +8
find the points where the curve meets the x-axis

Q2 gradient fucntion of the curve is given by 4x^3 -2x +3. the curve passes through (-1,2) where does the curve cut the y-axis

Q3 the point (2,-8) is a turning point of the curve y=f(x) f"(x) =12x +18
a) find the nature of the turning point at (2,-8)
b)find the equation of the curve

thanks..

 

[Mountain.Dew]

Q1 curve y=x^3 -2x^2 -4x +8
find the points where the curve meets the x-axis

Q2 gradient fucntion of the curve is given by 4x^3 -2x +3. the curve passes through (-1,2) where does the curve cut the y-axis

Q3 the point (2,-8) is a turning point of the curve y=f(x) f"(x) =12x +18
a) find the nature of the turning point at (2,-8)
b)find the equation of the curve

thanks..

im not going to do all the questions for you, but heres some handy tips:

1) questions asks for where the curve cuts the x-axis, IE solve the equation. start off by doing trial and error --> put in random numbers, like x = 1, x = 2...and see if any will come to zero. you now know one root --> so do long division to get a quadratic --> then its easy to find those roots.
2) for this question, u need to find the actual equation of the curve. you've been given its gradient, or DERIVATE, so to find the original equation, or the PRIMITIVE function, INTEGRATE ==> sub in pt (2,-8) to find c, ur constant, then sub x=0 to find where the curve cuts the y-axis.

3) (a) find nature of turning pt by finding f''(x). if f''(x) > 0, then the TP is MIN. IF f''(x) < 0, then the TP is MAX
(b)...there is one small problem...have u posted up the question correctly? there seems to be something missing with the function f'(x)...

 

[XcarvengerX]

Here's the answer. Check only if you have done it:

Spoiler

1. There are double roots at (2,0), which means the curve just touches the x-axis at this point. Another one is (-2,0), where the curve pass through this point.
2. The equation is y=x⁴ - x² +3x + 5. Therefore cut y-axis at (0,5)

Obviously there is a problem in question 3. I think you need a point for f'(x).

 

[sasquatch]

Q3 the point (2,-8) is a turning point of the curve y=f(x) f"(x) =12x +18
a) find the nature of the turning point at (2,-8)
b)find the equation of the curve


Doesnt seem to be a problem...

a) f''(x) = 12x + 18
f''(2) = 12(2) + 18
= 42
> 0

.:. point (2,-8) is a min turning point.

b) f''(x) = 12x + 18
f'(x) = 12x²/2 + 18x +c
= 6x² + 18x + c

When f'(x) = 0, x = 2

0 = 6(4) + 18(2) + c
c = -60

.:. f'(x) = 6x² + 18x - 60
f(x) = 6x³/3 + 18x²/2 - 60x + c
=2x³ + 9x² - 60x + c

Hmm now theres a problem..

 

[abcd9146]

dont you just shove (2,-8) into your f(x) and find c?, the point (2,-8) is on f(x).

you should get y=2x³+9x²-60x+60

 

[sasquatch]

OH YEAH hahaha...what the hell is wrong with me...

 

[XcarvengerX]

dont you just shove (2,-8) into your f(x) and find c?, the point (2,-8) is on f(x).

you should get y=2x³+9x²-60x+60

No, because f(x) isn't the same as f'(x) and so y isn't the same as y'.

 

[sasquatch]

No no he is right!...THe turning point (2,-8) still exists as a point on f(x). I.e. when x is 2, y is -8. So what abcd said is true.

TO make the question make more sense for you:

Q3 the point (2,-8) is a turning point of the curve y=f(x) f"(x) =12x +18
a) find the nature of the turning point at (2,-8)
b)find the equation of the curve

Meaning that the point (2,-8) exists as a turning point on the curve y = f(x). The second deriverative is given to help us determine the nature of the curve (for part a), and allow us to antidifferentiate to find both f'(x) and f(x).

When x = 2
f(x) = -8
f'(x) = 0

So this point can be used for both f(x), f'(x) to determine the constants arriving from the process of antidifferentiation.

 

[XcarvengerX]

OK. Better study more...

 

[rawker]

Oh my god. lol so much fighting!
I worked it out to be
f"(x) = 12x - 18
f'(x)= 6x^2 - 18x + C
when f'(x) = 0, x=2
0=6(4)+18(2)+C
C=-60
therefore f'(x)=6x^2 - 18x - 60
f(x) = 2x^3 - 9x^2 - 60x + C
(2,-8) -8=2(2)^3 - 9(2)^2 - 60(2) + C
C=60
f(x) = 2x^3 - 9x^2 - 60x + 60

 

[sasquatch]

Its not -9 its + 9

You posted previously f''(x) = 12x + 18

So if thats correct then it should be +9. But if you posted incorrectly (your previous post says - 18, then it'd be -9).

 

[rawker]

Its not -9 its + 9

You posted previously f''(x) = 12x + 18

So if thats correct then it should be +9. But if you posted incorrectly (your previous post says - 18, then it'd be -9).

True, posted incorrectly. Should be +9

 

[Mountain.Dew]

ummm one small problem...

if f'(x) = 12x + 18, then (2,-8) CANT be a turning pt...it should be when 12x + 18 = 0, when
x = -3/2...thats where the problem lies...

 

[sasquatch]

You read wrong!! f''(x) = 12x + 18

double dash = second deriverative!