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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y= x^3 - 3x^2 + 2
First we need to calculate the critical values of y, to do so, we need to find the...
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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.
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.
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.
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