differenciation?

Question

A curve has the equation: y=x^4 - 8x^3 + 16x^2 +8

a)Find dy/dx and d2y/dx2
b)Find the three values of x for which dy/dx = 0
c)Determine the coordinates of the point at which y has a maximum value

i'm having real problems with part B, through trial and error i managed to find two of the values for x (0 and 2) but the last one i can't seem to find, i would appreciate it if you would pote your working out also. Thanks!

Answer 1

dy/dx = 4x^3 - 24x^2 + 32x
4x(x^2 - 6x +8) = 0
4x(x - 2)(x - 4) = 0
so x can be 0, 2, 4

Answer 2

y = x^4 - 8x^3 + 16x^2 + 8

dy/dx = 4x^3 - 24x^2 + 32x

d^2y/dx^2 = 12x^2 - 48x + 32

b) Setting dy/dx = 0,

0 = 4x^3 - 24x^2 + 32x

Divide both sides by 4,

0 = x^3 - 6x^2 + 4x

Factor x out.

0 = x(x^2 - 6x + 4)

x = 0 and x^2 - 6x + 4 = 0

Solve the quadratic: x^2 - 6x + 9 + 4 = 9, (x - 3)^2 = 5,
x - 3 = +/- sqrt(5), x = 3 +/- sqrt(5)

Three values of x for which dy/dx = 0 are 0, 3 + sqrt(5), 3 - sqrt(5).

(c) To determine where y has a maximum value, all you have to do is plug in your critical numbers for your equation. If we call y = f(x), and
f(x) = x^4 - 8x^3 + 16x^2 + 8,

Just test
f(0), f(3+sqrt(5)), f(3 - sqrt(5))

And whatever you get as your highest value is your maximum.

Answer 3

a)
dy/dx = d( x^4 - 8x³ + 16x² +8 )/ dx
d( x^4 - 8x³ + 16x² +8 )/ dx = 4x³ - 24x² + 32x

d²y/dx² = d( 4x³ - 24x² + 32x )/ dx
d( 4x³ - 24x² + 32x )/ dx = 12x² - 48x + 32
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b)
4x³ - 24x² + 32x = 0
x³ - 6x² + 8x = 0
x² - 6x + 8 = 0

x = [6±(6²-4*1*8)^¹/2] / 2
x = [6±(36-32)^¹/2] / 2
x = [6±2] / 2

x' = 2
x'' = 4
x''' = 0
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c)
The maximum value is when dy/dx=0:
4x³ - 24x² + 32x = 0

=>
x' = 0
x'' = 2
x''' = 4

=>
Apply in x^4 - 8x³ + 16x² +8:
y' = 0^4 - 8*0³ + 16*0² +8 = 8
y'' = 2^4 - 8*2³ + 16*2² +8 = 16 - 64 + 64 + 8 = 24
y''' = 4^4 - 8*4³ + 16*4² +8 = 256 - 512 + 256 + 8 = 8

=>
The points are:
(0, 8), (2, 24) and (4, 8)

And the maximum point is:
p = (2, 24)

* If d²y/dx² is positive when dy/dx = 0, y is minimum;
* If d²y/dx² is negative when dy/dx = 0, y is maximum;
* If d²y/dx² = 0 when dy/dx = 0, there is no minimum nor maximum.

TEST:
when d²y/dx² = 0
12x² - 48x + 32 = 0
3x² - 12x + 8 = 0

The points are:
(0, 8), (2, 24) and (4, 8)

=>
y1 = 3*0² - 12*0 + 8 = 8
=> (0, 8) is minimum point

=>
y2 = 3*2² - 12*2 + 8 = 12 - 24 + 8 = -4
=> (2, 24) is maximum point

=>
y3 = 3*4² - 12*4 + 8 = 48 - 48 + 8 = 8
=> (4, 8) is minimum point

₢

Answer 4

Part A:

y = x^4 - 8x^3 + 16x^2 + 8
y' = 4x^3 - 24x^2 +32x (First derivative)
y'' = 12x^2 - 48x + 32 (Second derivative)

Part B:

Factor out 4x: 0 = 4x[x^2 - 6x + 8]
Reverse FOIL the inside of the right-side: 0 = (4x)(x-4)(x-2)
And we find that x={0,4,2}

Part C:

Set the first derivative equal to 0 to find a max/min, and solve. We found from Part B that the x-coordinate would be either 0, 4, or 2. Now we plug those values into the original equation to see which yields the highest value.

y(0) = 8
y(2) = 24
y(4) = 8

Thus at coordinate (8,24) we find the absolute max of the function.

Answer 5

A) Differentiating is simple, using the Power Chain Rule.

B) Differential 1 has the common factor 4x, and the remainder can be easily factored to yield roots at 4 and 2. Therefore the roots are 0, 4, and 2.

Answer 6

a) dy/dx= 4x`3 - 24x`2 + 32x
d2y/d2x= 12x`2 - 42x + 32

b) 4x`3- 24x`2 + 32x = 0
4x(x`2 -6x + 8) = 0
4x(x-4)(x-2) = 0
x = 0, 4, 2

c)use the first derivative test max at X=2

Answer 7

enable m signify what proportion cos[x subsequently 4 f cos4x could be cos mx Then for any value of m the spinoff is - msinmx subsequently - 4 sin 4x...........comparable as Robert's answer occasion spinoff of cos5x is - 5 sin 5x that's regularly shown employing chain rule For sin4x spinoff is 4 cos 4x

Answer 8

do u're h.w. ureself dude! thats called cheating! ;)