Absolute Maxima and Minima

[three3985]

Hello all,

I would like to check whether if my answers to the below questions are correct.






Thank you for taking the time to read my questions!

 

[Dr.Peterson]

You're right on the first problem. The absolute minimum is -2, at x=2.

For the second, check for sign errors! And is 0 in [1, 5]? Sketching a graph might help prevent those errors.

 

[R.M.]

On your first one, your answer of x = 2 is correct, however, your analysis has an error. You write that f ''(2) = 1, but f '' (x) is the horizontal line y = 4.

 

[three3985]

You're right on the first problem. The absolute minimum is -2, at x=2.

For the second, check for sign errors! And is 0 in [1, 5]? Sketching a graph might help prevent those errors.

Just realised I've made some error. So to clarify, the absolute minimum for A is 0 whilst the absolute minimum for B is 1?

 

[three3985]

On your first one, your answer of x = 2 is correct, however, your analysis has an error. You write that f ''(2) = 1, but f '' (x) is the horizontal line y = 4.

Hey, thanks for replying, I wrote f ''(2) = 1>0, f (2) = -2 would be the absolute minimum, is that incorrect?

 

[Dr.Peterson]

Just realised I've made some error. So to clarify, the absolute minimum for A is 0 whilst the absolute minimum for B is 1?

You meant maximum there, right?

Hey, thanks for replying, I wrote f ''(2) = 1>0, f (2) = -2 would be the absolute minimum, is that incorrect?

You had previously written, f"(x) = 4, so clearly f"(2) = 4, not 1, right? Where did you get 1?

 

[three3985]

You meant maximum there, right?

You had previously written, f"(x) = 4, so clearly f"(2) = 4, not 1, right? Where did you get 1?

1) Isn't the maximum 25 and the minimum 1 for B?

2) Oh thanks for spotting that out for me, I really appreciate it.

 

[Dr.Peterson]

1) Isn't the maximum 25 and the minimum 1 for B?

I misread what you wrote, thinking "the absolute minimum for A is 0 whilst the absolute minimum for B is 1" was all about A (the minimum and maximum). I missed "for B".

 

[HallsofIvy]

Personally, on neither of those would I take the derivative. Instead, complete the square!

$2x^2- 8x+ 6= 2(x^2- 4x)+ 6= 2(x^2- 4x+ 4- 4)+ 6= 2(x- 2)^2- 8$.
Since a square is never the minimum is -8 when x= 2. We also need to check the endpoint x= 0. $2(0^2)- 8(0)= 6= 6$ and as x goes to infinity, the value increases indefinitely.

The absolute minimum is -8 and there is no absolute maximum.

$x^2$ is already a perfect square. Its minimum is 0 at x= 0. Then we need to check the endpoints of the intervals.

A. [-1, 1] $(-1)^2= 1$, and $1^2= 1$, The interval includes x= 0 so the minimum is 0 and the maximum is 1.

B. [1, 5] $1^2= 1$ and $5^2= 25$. The minimum is 1 and the maximum is 25.

C. [-5, 5] $(-5)^2= 26$ and $5^2= 25$. The interval includes 0 so the minimum is 0 and the maximum is 25.