math question solving problem

Question

Let f(x)=x^2−8x+13. 

Find the critical point c of f(x) and compute f(c).

The critical point c is =  

The value of f(c) =  

Compute the value of f(x) at the endpoints of the interval [0,8].

f(0) =  

f(8) =  

Determine the min and max of f(x) on [0,8].

Minimum value =  

Maximum value =  

Find the extreme values of f(x) on [0,1].

Minimum value =  

Maximum value =  

Answer 1

To find the critical point, take the derivative and set it equal to zero, then solve. That solution (x = __) will be the value of c.

For f(x) = x² - 8x + 13, use the power rule to take the derivative. f '(x) = 2x - 8

To find the critical points, make f '(x) = 0

2x - 8 = 0

2x = 8

x = 4

So c (the critical point) = 4

The function value at c = 4 or f(c): f(4) = (4)² - 8(4) + 13 = 16 - 32 + 13 = -3

For the interval [0, 8] the function values at the endpoints are:

f(0) = (0)² - 8(0) + 13 = 0 - 0 + 13 = 13

f(8) = (8)² - 8(8) + 13 = 64 - 64 + 13 = 13

The minimum on [0, 8] occurs at x = 4 (the critical point) and is -3

The maximum on [0, 8] occurs at x = 0 and x = 8 (the end points) and is 13

For the interval [0, 1]:

There are no critical points on [0,1] because the only critical point was at x = 4. So the find the minimum and maximum on the interval, we just need to check the endpoints:

f(0) = (0)² - 8(0) + 13 = 0 - 0 + 13 = 13

f(1) = (1)² - 8(1) + 13 = 1 - 8 + 13 = 6

The minimum on [0, 1] occurs at x = 1 (an endpoint) and is 6

The maximum on [0, 1] occurs at x = 0 (an endpoint) and is 13

I hope this helps.