Differentiation application

Question

f(x)=x^4-4x^3+4x^2+1 

what is the maximum and minimum of f(x) on the interval [-1,5]

Answer 1

The critical values are

0 =f'(X) = 4x^3 - 12x^2 + 8x

 

0 = x(4x^2 - 12 x + 8)

0 = x(x^2 - 3x + 2)

0 = x(x      -   2)(x     -        1)

 

x=0, 1, 2

 

 

f(-1) = (-1)^4 - 4(-1)^3 + 4(-1)^2 + 1

         =  1  + 4 + 4 +1 

         = 10

 

 

f(0) = 1

f(1) = 1 - 4 + 4 + 1 = 2

f(2) = 2^4 - 4*2^3 + 4*2^2 +1

       = 16 - 32 + 16 + 1

       = 1

f(5) = 5^4 - 4*5^3 + 4*5^2 + 1

      =  625 - 500 + 100 + 1

     = 226

 

max is 226

min is 1

 

 

 

Answer 2

Hi, Nhu.

 

To find the minimum or maximum in Calculus, you find the derivative, then set the derivative = 0, the x's that cause the derivative to = 0 will be your possible minimums and maximum.

 

f'(x) (the deriv) = 4x³ - 12x² + 8x

 

Factor this out and get that f'(x) = 4x(x² - 3x + 2)  or

 

f'(x) = 4x(x-2)(x-1)

 

Setting this to 0, get 4x(x-2)(x-1)=0 so the critical numbers (the x's that cause the derivative to be = 0) are

 

x = 0,  x = 1,  and x = 2.

 

If you draw a sign line with these three critical numbers, you will find that the derivative has these characteristics:

 

   -   |   +   |   -   |   +

----------------------------

       0        1       2

 

Because the deriv changes from - to + at x=0, f (the original function) changes from decreasing to increasing at x=0  (like a "U"), so there is a minimum at x = 0.

 

Because the deriv changes from + to  - at x=1, the orig function changes from increasing to decreasing at x=1 (like "^" but rounded), so there is a maximum at x = 1.

 

Because the deriv changes from - to + at x=2, the orig function changes from decreasing to increasing at x=2 (like a "U"), so there is a minimum at x = 2.

 

BUT the problem did not ask for the location of the minimums and maximums (the x-coordinates), it asked for the value of the minimum and maximum.  

 

So there is a relative min at (0,1) and at (2,1).  There is a relative max at (1,2).

 

The relative min is 1 (y-value of both mins).  The relative max is 2 (y-value of the max).

 

BUT...  the teacher gave you a CLOSED interval to check, so besides these relative min and max, he/she may want the absolute min and max on that closed interval.  In that case, we also need to check the endpoints of the interval.

 

The interval is [-1,5].  So need to find f(-1)  (x=-1 plugged into the ORIGINAL function, f)

Also need to find f(5)  (x=5 plugged into the ORIGINAL function, f, NOT the derivative f')

 

f(-1) = 10           f(5) = 226

 

If you must give the absolute min on [-1,5], it is 1 since this is the smallest y-value we know of in [-1, 5]

But if you must give the ABSOLUTE max on [-1,5], it is located at the endpoint of the interval, at x = 5 because the y-value (maximum value) at this end of the interval is 226 which is way more than the relative max of 2.

 

So...

 

If teacher wants relative min (1)  and relative max (2) these are the correct answers.

If teacher wants absolute min (1) and absolute max (226) these are the correct answer.