quick calculus question

[thirtythree]

for the function f, f''(x)=24x-18 and f'(x)=-6. how is this possible? maybe it's just late, but it doesn't seem like the slope of f could be constant if the slope of the slope isn't. it's on an old ap question so I can't imagine it's a typo.

 

[waffel]

Yeah, that doesn't make any sense. If you take the derivative of f'(x) to get f''(x) you get f''(x) = 0.
Could it be that they meant f'(x0) = -6? (x0 is x-sub-zero)

 

[thirtythree]

Originally posted by: waffel
Yeah, that doesn't make any sense. If you take the derivative of f'(x) to get f''(x) you get f''(x) = 0.
Could it be that they meant f'(x0) = -6? (x0 is x-sub-zero)

it's just an x. the questions don't make sense given the info either. e.g. find each x such that the line tangent to the graph of f at (x,f(x)) is horizontal. the slope is always -6 though

 

[waffel]

Originally posted by: thirtythree

Originally posted by: waffel
Yeah, that doesn't make any sense. If you take the derivative of f'(x) to get f''(x) you get f''(x) = 0.
Could it be that they meant f'(x0) = -6? (x0 is x-sub-zero)

it's just an x. the questions don't make sense given the info either. e.g. find each x such that the line tangent to the graph of f at (x,f(x)) is horizontal. the slope is always -6 though

Well that doesn't make a damn bit of sense
(But it's been a few years since I took the AP test, and it's late)

 

[misle]

Well, if you integrate f''(x) you get f'(x) = 12 x^2 - 18x

If x=1 then f'(x) = 12-18 = -6

Seems like an odd problem though.

 

[bonkers325]

its not possible unless you have some sort of initial or bounding conditions

 

[thirtythree]

Originally posted by: misle
Well, if you integrate f''(x) you get f'(x) = 12 x^2 - 18x

If x=1 then f'(x) = 12-18 = -6

Seems like an odd problem though.

when you integrate f''(x) you get 12x^2-18x+c, so f'(1) doesn't necessarily equal -6. perhaps my teacher retyped the question rather than copying it directly and made some sort of mistake in the process. I'll ask him about it today.

 

[misle]

Originally posted by: thirtythree

Originally posted by: misle
Well, if you integrate f''(x) you get f'(x) = 12 x^2 - 18x

If x=1 then f'(x) = 12-18 = -6

Seems like an odd problem though.

when you integrate f''(x) you get 12x^2-18x+c, so f'(1) doesn't necessarily equal -6. perhaps my teacher retyped the question rather than copying it directly and made some sort of mistake in the process. I'll ask him about it today.

Yes, that's correct... damn C's... it's early, I'm tired, insert other lame excuse here.