Find the maximum value of f(x)= x2-2x+1 on the interval [0,3]

Question

Calculus AB

Answer 1

First, we need to find the critical points of the function, which we can do by finding the derivative of the function:

 

f(x)= x²-2x+1 so the derivative is f'(x)= 2x-2

 

Now find f'(0), which is: 0= 2x-2 or x=1.

 

The next step is evaluate the original function with both the critical point and the interval points 0 and 3.

 

f(0)= (0)²-2(0)+1= 1

f(1)= (1)²-2(1)+1= 0

f(3)= (3)²-2(3)+1= 4

 

So based upon this information, the maximum value for this function on that interval will be 4. 

Answer 2

Set the derivative of f(x) equal to zero.

 

2x - 2 = 0

 

 

Solve for x.

 

2(x - 1) = 0

 

x = 1

 

Since f(x) is a parabola that opens upward, f(1) is a local minimum.  But could also be the maximum value.  So we would have to compare it to the other values at the end boundaries.

 

Now, we evaluate f(x) at the end boundaries.

 

f(0) = 1

f(1) = 1 - 2 + 1 = 0

f(3) = 9 - 6 + 1 = 4

 

The greatest value is 4.  Therefore, 4 is the maximum value.