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/sci/ - Science & Math

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3248207 No.3248207 [Reply] [Original]

/sci/ I'd really appreciate your help here. It's a stupid question but I can't figure out why the answer is 25/4:

f(x) = x^2 - x + 2 is on the interval [-2, 2]. If M is the absolute max value on this interval and m is the absolute min, what is M - m?

I know, I'm an idiot.

>> No.3248229

Bumping so I don't become more of a failure at life.

>> No.3248236

graph that function
???
profit

>> No.3248242

here is a hint... Find the absolute min value.
1.) Take derivative of f(x) = x^2 -x +2
2.) Solve for x to find either the max or min
3.) Plug the answer for x of the derivative back into the original function to find the y value
4.)???????
5.) Profit

>> No.3248250

Find critical point

Find value of function at critical point (7/4)

Find value of function at each end point (4 and 8)

Min/Max is at critical point or endpoint

Min is 7/4, Max is 8

32/4 - 7/4 = 25/4

>> No.3248251

>>3248207
>>3248207
Answer is 1. Not even joking.

>> No.3248259

>>3248251

I know, that's what I got. 1 is not even a possible answer on the multiple choice. I'm pulling this question from a past test in the course - the university is reputable and I doubt they'd be THAT wrong about this whole thing. There must be a way for it to be 25/4.

>> No.3248267

>>3248250

I totally missed this answer. That makes perfect sense, thank you.

>> No.3248272

>>3248250

Actually, how did you get the critical point 7/4?

I get 2x - 1 = 2(x - 1/2) => critical pt @ 1/2.

>> No.3248278

>>3248259
>>3248259
....I was kidding.

>Take derivative of f(x) = x^2 - x + 2
>Derivative: f(x) = 2x -1
>Solve for x: x= 1/2
>Positive, so minimum. Plug back into f(x) = x^2 - x + 2 to find the critical point of (7/4)
> Plug the upper and lower ranges into your function to find that f(-2) = 8
> M = 8, m = 7/4
> M - m = 25/4
>????????
>Profit

>> No.3248279

>>3248272

That's the first step, now do the second step, i.e plug it in.

>> No.3248306

So I suppose all my problems came from a poor understanding of the definition of a critical point: When I solve for x = 1/2, is that already a critical point, or is it only once I plug it into the original function that it is defined as a critical point? What is actually going on behind the scenes in this procedure? I've found the x value for a point where the tangent is horizontal (a critical point) and then plug that x value into the original function to determine the y value that exists at the point where the tangent is horizontal? Is that right?

>> No.3248346

>>3248306

You find the critical number which is an x value. Maximums and minimums are the y values at that particular x value.

The x and y value together is the critical point.

>> No.3248350

>>3248346

Nevermind, the critical point is strictly the x value and the y value is called the critical value.

>> No.3248370

>>3248350

Thanks

>> No.3248429

Post the rest of the spark charts thing.