@MileSwimmerDude can u explain why its not that first deriv cant = 0
i got the first derivative to be always positive
Yeah f’ had to be > 0 because a and b could still get infinitely close to each other. At least that’s how I thought of it.
I don’t even know what a Taylor Series is I’m guessing it’s something you learn in BC which I’m taking next year.
oh thats so weird so they would be so close that subtracting them would be essentially 0 when getting the slope?
@nhljohn871 because it can have a ‘hesitation’ point where it’s limit is zero, but cant go negative.
@nhljohn871 It’s because it is saying what MUST be true. If it equals 0 at a single point that doesn’t effect anything because the function still stays monotonic as long as it goes back to increasing after the mtan is zero at that one point.
For the f(a) f(b) one I’m pretty sure that it was it has to pass through 10 twice bc it said it was continuous and even though a is always > b and f(a) is always > f(b) there could be a point where the slope is zero like on x^3 and it would still work out
ok that makes sense thanks
I put D. Cant equal zero. The other choice A said greater than or equal to zero if u read properly so A is wrong.
Yeah I got it had to pass through 10 twice. But I think that was another question entirely.
@javacash It can equal zero.
I got that to, f(x)=10 twice
@Javacash01 Consider f(x)=x^3. f(a)<f(b) for every a<b, but f’(0)=0.
Nope! It cant cause in the interval for every single combination b was always greater than a so there would never be a derivative equal to zero.
Saying that the derivative is zero means that f(b) = f(a) which violates what was stated in the problem.
So is it that f prime doesn’t equal zero
@Javacash01 not necessarily because f’(0)=0 only at the one point (0,0)
@Javacash01 It doesn’t matter. f(a) and f(b) can get infinitely close to each other in value BUT they can never equal each other. If the slope of the tangent line is 0 at a single point f(b) is still > f(a).
The questions domain did not include 0
approximation of f’(16)=5?
A was greater than zero