@calcstu I was wondering the same thing. I said there was no point of inflection.
The second derivative went from 0 to still 0 so not point of inflection at x=4. Sorry for me english.
@MutantsAbyss hmmm… does a sharp point still count as a max? Because if it is, then there is a POI…
@LeoB1729 @calcstu because f goes from increasing to decreasing, then g’’(x) would have had to go from positive to negative which means that g went from concave up to concave down I think. that was my thought process
@writerzt71 Oh yeah… that makes sense. Oops. Did anyone get -8 and 8 for 3c?
Critical points can occur the next immediate derivative of the function in question equals 0 or undefined.
Ok so how do you do 2 C/D and 4B?
Nevermind I just saw that previous post, I think I didn’t do too badly!!
how to do number 3 C
i’m waiting for mr calculus on 2c/d. other people apparently got the intervals I got for 2d but I don’t know if I’d get credit because I provided no work
How did people get -8 and 8 for 3c? The interval only goes from -4 to 12. Also, for 3d, isn’t it just [-4,-2] and [6,12] cuz g(x) is less than 0 when area under f(t) is negative?
The CB seems create some new-style FRQ compared with previous years. Like the funnel and parametric ones
Mr. Calculus answers coming soon…
@SATsRuinLife they are asking for the absolute min & max VALUE, not the values of x that give the min/max.
Here’s my attempt on No. 3:
3a
x = 10 is neither a min or max, because g’(x) = f(x) does not change sign there.
3b
x = 4 is POI for g b/c g"(x) = f '(x) changes sign there.
3c
check where g’(x) = f(x) = 0, which are –2, 2, 6, 10
also check endpoints –4, 12
g(–4) = ∫ (from 2 to –4) f(t) dt = –(–4 + 8) = –4
g(–2) = ∫ (from 2 to –2) f(t) dt = –(8) = –8 <-- absolute min value
g(2) = ∫ (from 2 to 2) f(t) dt = 0
g(6) = ∫ (from 2 to 6) f(t) dt = 8 <-- absolute max value
g(10) = ∫ (from 2 to 10) f(t) dt = 8 – 8 = 0
g(12) = ∫ (from 2 to 12) f(t) dt = 8 – 8 – 4 = –4
3d
g(x) <= 0 on [–4, 2] and [10, 12]
My attempt at No. 6:
6a
f(x) = 1 – (1/2)(x – 1) + (1/8)(x – 1)^2 – (1/24)(x – 1)^3 + … + [(–1)^n/(n*2^n)] * (x – 1)^n + …
Note: general term: [f^(n)(1) / n!] * (x – 1)^n, which simplifies to [(–1)^n/(n*2^n)] * (x – 1)^n
6b
Interval of convergence would be from –1 to 3, but need to check endpoints
x = –1:
sum [(-1)^n(-2)^n over n2^n] = sum (1/n)
harmonic series, so it diverges
x = 3:
sum [(-1)^n(2)^n over n2^n] = sum [(-1)^n/n]
alternating harmonic series, so it converges
Therefore, interval of convergence is (–1, 3]
6c
f(1.2) ≈ 1 – (1/2)(0.2) + (1/8)(0.2)^2 = 0.905
6d
alternating series remainder <= first neglected term, so
error <= (1/24)(0.2)^3 = 1/3000 < 0.001
Finally, my attempt at No. 2:
[Note: x’(t) = t^2 + sin(3t^2)]
2a
x(3) = 5 + ∫ (from 0 to 3) x’(t) dt = 14.377
y(3) = –1/2
position of particle at t = 3: (14.377, –1/2)
2b
y’(3) = 1/2
x’(3) = 9.956
dy/dx (at t = 3) = y’(3)/x’(3) = 0.050
2c
speed (@ t = 3) = sqrt[(x’(3))^2 + (y’(3))^2] = 9.969
2d
total distance = ∫ (from 0 to 1) sqrt[(x’(t))^2 + (y’(t))^2] dt + ∫ (from 1 to 2) x’(t) dt = 4.350
(b/c the particle is not moving vertically from t = 1 to t = 2)
For 2d, I split it up into 3 intervals to find the total distance.
The particle is still moving on the interval 1 to 2 (in the x-direction), so you would still have to find the distance traveled.
FUNNEL FLASHBACK: