find the values of x where the function:

y= x^3 -3x^2 + 2 reaches a max and min, also calculate the max & min values, sketch the function

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Given:

f(x) = y = x^3 - 3x^2 + 2

To fin value of x where value of y reaches max and min we must find f'(x) an equate it to 0. Thus:

f'(x) = 3x^2 + 3*2x = 3x^2 + 6x

Equating f'(x) to 0:

3x^2 + 6x = 0

==> 3x(x + 2) = 0

Thus x = 0 or x = -2

Substituting these values of x in the f(x) we will get max and min value of y.

f(0) = 0^3 - 3*0^2 + 2

= 2

f(-2) = (-2)^3 - 3(-2)^2 + 2

= -8 - 3*4 + 2 = -8 - 12 + 2 = -18

Thus:

Maximum value of y = 2. This occurs when x = 0.

Minimum value of y = -18. This occurs when x = -2.

y= x^3 - 3x^2 + 2

First we need to calculate the critical values of y, to do so, we need to find the first derivative's zeros.

Let us differentiate:

y' = 3x^2 - 6x

3x^2 - 6x = 0

Factor 3x:

==> 3x(x-2) = 0

The, critical values are x= {0, 2}

when x= 0==> y= 2

when x= 2 ==> y= -2

Then extreme values are: (0, 2) and (2, -2)

We have the given function y=x^3-3x^2+2.

To find the values of x where the function reaches a maximum and minimum value we need to differentiate y and equate that to 0.

Differentiating y, we get y'=3x^2-6x.

Equating it to 0, 3x^2-6x=0 =>3x(x-2)=0 => x= 0 or 2.

Now differentiate y'. We get y''=6x-6

For x=0, 6x-6=0-6=-6. As it is negative we have a maximum value at x=0 and it is x^3-3x^2+2= 2.

For x=2, 6x-6=12-6=6. As it is positive we have a minimum value at x=2 and it is x^3-3x^2+2= 8-12+2= -2.

Therefore for y=x^3-3x^2+2,

**The minimum value is -2, reached when x=2.**

**The maximum value is 2, reached when x=0. **

y= x^3-3x^2+2

To find maximum and mimimum and trace the graph.

Solution:

We find the critical points x1 and x2 by setting f'(x) = 0 and solve for x. We find the the sign of f"(x) at these critical points. If f " (x) is negative (or positive )at the critical point , then f(x) should be maximum (or minimum) respectively.

f'(x) = (x^3-3x^2+2)' = 3x^2-3*2x =3x^2-6x

f'(x) = 0 gives 3x^2-6x = 0. 3x(x-2) = 0

So x= 0 Or x =2 are the critical values.

Now find the 2nd derivative.

f''(x) = (3x^2-6x)' = 6x-6.

Find the sign of f"(x) at the critical points x = 0 and x = 2.

f '' (0) =6*0 -6 = -6. S0 f(0) is a local maximum at x= 0. f(0) = 0^3-3*0^2+2 = 2

So f(0) = 2 is a maximum.

At x= 2, f " (2) = 6*2-6 = 12-6 = +6. So f(2) is the local minimum.

f(2) = 2^3-3*2^2+2 = 8-12+2 = -2 is the local minimum.

Tracing the graph:

Roots:

Obviously f(1) = 0 . So x=1 is a real root of x^3-3x^2+2 = 0. The graph crosses x axis at x=1.

Also (x^3-3x^2+2)/(x-1) = x^2-2x-1 becomes 0 for x^2-2x+1 =2. Or

(x-1)^2 =2. Therefore,

x-1 = +or- sqrt2 .Or

x1 = 1+sqrt2 and x2 = 1-sqrt2.

So f(x) has three real roots : A pair of surds and one rational . x1 = 1-sqrt2, x=1 and x2 = 1+sqrt2.

So f(x) crosses x axis 3 times.

y = 2 is the y intercept wnen x = 0. Also it is the local maximum.

Graph and its nature:

At -infinity, f(x) = -infinity as the eading term is x^3 has a positive coefficient 1.

The graph of f(x) is strictly increasing for all x < 0 as f'(x) =3x(x-2) is positive for x<2.. So f(x ) is continuously increasing from - infinity to a maximum in the interval (-infinity to 0). It crosses x axis at x 1 = 1-sqrt2.

It reaches a local maximim at of f(0) = 2 at x =0.

In the interval (0 , 2), f(x) continuously decreasing from maximum local maximum 2 to -2. The graoh crosses x axis at x=1 as f(1) = 0.

At x=2 , f(x) reaches a local minimum off(2) = -2.

For x>2 as f'(x) = 3x(x-2) > 0. So f(x) is strictly increasing again. The graph crosses x axis at x2 = 1+sqrt2 and f(x) goes infinity as x--> infinity.

Since f'(x) = 0 at x=0 and x =2, the graph has tangent at x = 0 and at x= 2 || to x axis.

The shape of graph is like that of a curved letter "N " in the interval (1-sqrt2 , 1+sqrt2), further, with left extremity approching - infinity as x-->-infinity and right extremity going to +infinity as x--> +infinity.

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