Turning point questions!! (1 Viewer)

[Seraph]

ok i have a few questions , i dont seem to be getting these right

my first question is
Find the turning points on the curve y=(4x^2 - 1)^4

okay so we use the chain rule yes?

first derivatve
y' = 4(4x^2 - 1)3 x 8x
so y' = 32x(4x^2 - 1)
put that to 0 to find points
32x(4x^2 - 1)^3 = 0
so um 4x^2 = 1
x^2 = 1/4
so x = 0.5
x = -0.5
and x = 0
sub into y
(0.5,0)
(-0.5,0)
(0,1)
Okay now at this point i got my second derivative
now my question here is do i use the chain rule??? if so then i end up with
768x^2 (4x^2 - 1 )^2 yes?
okay i used the (0.5,0) and (-0.5,0) at first they were both possible points of inflection but they turned out to be both minimum points.... however the point (0,1) is supposed to be a maximum point however i keep getting it as a minimum


my next question is for y=(x-3)(4-x)^1/2
how do i differentiate this???
(first and second) ?

 

[spice girl]

imho, the massive differentiation is to tell your teacher you know what you're doing, but in practice you can '*usually* work out where the turning points are by using graphing techniques.

e.g. y=(4x^2 - 1)^4

first consider the graph y = 4x^2 - 1
4 regions of interest: x < -1, -1 < x < 0, 0 < x < 1, x > 1
(for convenience i'll refer to these as regions 1, 2, 3, 4 respectively)

at region 1: y is positive and decreasing
at region 2: y is negative and decreasing
at region 3: y is negative and increasing
at region 4: y is positive and decreasing

we know that given a > b > 0, then a^4 > b^4, but when c < d < 0, d^4 < c^4

thus, applying these properties to draw y=(4x^2 - 1)^4 from the initial graph y = 4x^2 - 1, we see local minimums between region 1 and 2, and between 3 and 4 (i.e. x = -1, +1), and a local maximum between region 2 and 3 (i.e. x = 0). Also by the above properties, we conclude there can be no other turning points.

 

[Seraph]

hmm i sort of understand that , but i really want to know how to do it via differentation method thx anyway

 

[...]

Originally posted by Seraph
but i really want to know how to do it via differentation method thx anyway

the second derivate?

 

[Seraph]

yea i know how to use it .. its just for those questions... i wasnt getting a right answer on point (0,1) for that first question


and the 2nd one was how on earth do i differentiate this (1st and 2nd)
y=(x-3)(4-x)^1/2

 

[Xayma]

Originally posted by Seraph
yea i know how to use it .. its just for those questions... i wasnt getting a right answer on point (0,1) for that first question


and the 2nd one was how on earth do i differentiate this (1st and 2nd)
y=(x-3)(4-x)^1/2

Product Rule, x2. ie. dy/dx=v(du/dx)+u(dv/dx) so that will become
(4-x)^1/2.1+(x-3).1/2(4-x)^(-1/2).-1=(4-x)^1/2-(1/2)(x-3)(4-x)^(-1/2) and then the product rule again for the second part of that, for the first half just use the chain rule.

 

[Seraph]

yikes this is the answer i got..
so how would i find the turning points and nature of this

y'' = (x-4)^1/2.1/4 (x-3)(4-x)^-1/5 - 1/2 (x-3)(4-x)^-1/2 . 1/2(x-4)^-1/2

 

[zeropoint]

Originally posted by Seraph
768x^2 (4x^2 - 1 )^2 yes?

No.
y = f(x)
f(x) = (4x^2 − 1)^4
df/dx = 32x (4x^2 − 1)^3
d^2f/dx^2 = 768x (4x^2 − 1)^2 + 32 (4x^2 − 1)^3 (product rule)
f''(0) < 0.
Therefore (0,1) is a local maximum.
f''(−0.5) = f''(0.5) = 0
Therefore no conclusion can be drawn. Test the sign of the gradient on each side of the critical points.
f'(−0.5 − epsilon) < 0
f'(−0.5 + epsilon) > 0
Therefore (−0.5,0) is a local minimum.
f'(0.5 − epsilon) < 0
f'(0.5 + epsilon) > 0
Therefore (0.5,0) is a local minimum.

Originally posted by Seraph
yikes this is the answer i got..
so how would i find the turning points and nature of this

y'' = (x-4)^1/2.1/4 (x-3)(4-x)^-1/5 - 1/2 (x-3)(4-x)^-1/2 . 1/2(x-4)^-1/2

Set df/dx = 0
(4 − x)^0.5 + (3 − x)/(2(4 − x)^0.5) = 0

Multiply both sides by (4 − x)^0.5

4 − x + 3/2 − x/2 = 0
8 − 2x + 3 − x = 0
x = 11/3

d^2f/dx^2 = −(4 − x)^−0.5 + (3 − x)/(4(4 − x)^3/2)
f''(11/3) < 0.

Therefore (11/3 , 2/(3(3)^0.5)) is a local maximum.