Post by: Bestie on May 06, 2014, 09:59:00 pm
Thank you in advance
Post by: RKTR on May 06, 2014, 10:30:31 pm
range of cos x =[-1,1]
range of cos^2(x) =[0,1]
range of 1/cos^2(x) = [1, infinity)
therefore it is always positive
Post by: Bestie on May 07, 2014, 11:36:40 am
also, can you please help me here:
how would i simplify (50/4)cos(x)^2 + (18/4)sin(x)^2 + 8 cause apparently it needs to equal 5?
Post by: kinslayer on May 07, 2014, 01:56:15 pm
oh yea how stupid of me thank you :)
also, can you please help me here:
how would i simplify (50/4)cos(x)^2 + (18/4)sin(x)^2 + 8 cause apparently it needs to equal 5?
This never equals 5...
Something missing from the question, perhaps?
Post by: keltingmeith on May 07, 2014, 09:36:26 pm
Post by: lzxnl on May 07, 2014, 10:36:00 pm
You can just say the square of a real number is always positive. sec^2 (x) is the square of a real number as sec(x) returns real values for real values of x.
Post by: Bestie on May 24, 2014, 08:59:22 pm
question:
how would i find the coordinates of the inflexion of y = sin(x) for x [0,2pi]
what i have done so far:
f'(x) = cos(x)
f''(x) = -sin(x)
f''(x) = 0 , x = pi/2 and 3pi/2
i'm lost what now?
thank you in advance
Post by: RKTR on May 24, 2014, 09:06:51 pm
-sin(x)=0
x=0,pi,2pi i think you did f'(x)=0 instead?
sub the values back into y=sin(x)
the coordinates are (0,0),(pi,0),(2pi,0)
Post by: Bestie on May 24, 2014, 09:09:37 pm
and yup i accidentally did f'(x) = 0
sorry
Post by: Thorium on May 24, 2014, 09:26:56 pm
but then why is the ans only (pi,0) ?
and yup i accidentally did f'(x) = 0
sorry
It has to be only (pi,0) because x is restricted to [0,2pi]. I accept that 0 and 2pi are still included, but the points of inflexions at those points are not shown conpletely if you plot them. In contrast, the point of inflexion at (pi,0) is drawn completely.
Post by: Bestie on May 24, 2014, 09:43:26 pm
Post by: EspoirTron on May 24, 2014, 09:50:37 pm
oh... cause the ans used something about proving that f''(x) is a minimum for it to be a point of inflection, but shouldn't f''(x) = 0
Use a concavity test for f''(x)=0. If concavity changes then it is an inflection point, but not necessarily a stationary point of inflection.
Edit: I was confused.
Post by: Bestie on May 24, 2014, 10:00:12 pm
Post by: Thorium on May 24, 2014, 10:08:19 pm
im confused too, cause then there is no need to do f''(x) cause you know there is an inflection already???
There is a point of inflexion at x if and only if f"(x)=0 at x. Since f"(x) is the derivative function of f'(x), then we can say that a point of inflexion in f(x) is a local min/max in the f'(x).
Post by: Bestie on May 24, 2014, 10:12:19 pm
Post by: EspoirTron on May 24, 2014, 11:09:17 pm
There is a point of inflexion at x if and only if f"(x)=0 at x. Since f"(x) is the derivative function of f'(x), then we can say that a point of inflexion in f(x) is a local min/max in the f'(x).
I would be very careful in making that assumption. Consider f(x)=x^4.
Post by: lzxnl on May 25, 2014, 12:19:21 am
There is a point of inflexion at x if and only if f"(x)=0 at x. Since f"(x) is the derivative function of f'(x), then we can say that a point of inflexion in f(x) is a local min/max in the f'(x).
There is a point of inflection at x=a if and only if f''(x) changes sign at x=a (and, of course, f(x) is twice differentiable at x=a).
So, what you can say is that if f''(a)=0 and f'''(a) is not 0, x=a is a point of inflection. If f'''(a)=0, you'll need to examine higher order derivatives; if you keep differentiating and you keep getting zeros at x=a, if the first non-zero derivative after the second derivative is an even derivative, you have a stationary point (classified by the sign of this even derivative; positive means local minimum, negative means local maximum) and if the first non-zero derivative is an odd derivative, you have a stationary point of inflection. This assumes all higher order derivatives exist at x=a. Confused? Well you generally won't come across functions in VCE with first, second and third derivatives all zero.
Post by: Bestie on May 31, 2014, 02:50:53 pm
Thank you
Mod Edit: Took out question that was answered elsewhere - Phy124