Determining Extrema for f(x) = x^2 - 8x + 9

[scorpa]

Hello!

I am doing a question about curve sketching and determining extrema and just need some clarification. Here is the question:

Determine the extreme values of the function f(x) = x^2 - 8x + 9, where
-1 is greater than or equal to x is less than or equal to 5.

This is what I have done so far:

I decided to check for stationary points, as extreme values may occur at f'(0)=0.

f'(x) = 2x-8
2x-8 = 0
x = 4

f(4) = 4^2 - 8(4) +9 = -7
(4,-7) may be a max, min or neither.

f(3.9) = (3.9)^2 - 8(3.9)+9 = -6.99
f(4.1) = (4.1)^2 - 8(4.1)+9 = -6.99

Does this mean that this point cannot be used in determining the absolute minimum? I am totally lost at this point. I was going to use the endpoints given to figure out the maximum, but I'm not sure if that is an acceptable method. Thanks for any help you can give me.

 

[codyg1985]

You can use the endpoints of the interval as well as x = 4. You are allowed to use the endpoints because they are included in the interval.

 

[dextercioby]

The function is continuous on the whole domain.It's a parabola.The theory says that the sign of "a" (=1,in this case) decides the nature of the extremum...In this case,it's a minimum...

That's all there's to it.

Daniel.