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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: username on June 19, 2008, 04:59:21 pm

Title: Differentiation help please?
Post by: username on June 19, 2008, 04:59:21 pm: Hellooo. I've got 2 questions I really need help with, so if anyone could, thanks!

The graph of the function with rule $\frac{k}{2^{3}+1}$ has gradient 1 when x = 1. The value of K is?

The equation 3sinx-1=b, where b is a positive real number, has one solution in the interval (0, 2pi). The value of b is?
Title: Re: Differentiation help please?
Post by: Mao on June 19, 2008, 05:04:06 pm: erm... why is it $\frac{k}{2^3+1}$ ? did you mean $\frac{k}{2^x+1}$ or $\frac{k}{x^3+1}$ ?

for the second one, recognise that the interval represent a full period => sine is either at its maximum or minimum.
Title: Re: Differentiation help please?
Post by: username on June 19, 2008, 05:30:23 pm: Sorry I'm new to the coding thing. It's k / (2(x^3)+1)

u_u
Title: Re: Differentiation help please?
Post by: Toothpaste on June 19, 2008, 05:39:09 pm: Quote from: username on June 19, 2008, 05:30:23 pm
Sorry I'm new to the coding thing. It's k / (2(x^3)+1)

u_u

$\frac{d}{dx}(\frac{k}{(2x^3+1)})= \frac{-6x^2k}{(2x^3+1)^2}$

$x=1, \frac{d}{dx}=1$

$\frac{-6(1)^2k}{(2(1)^3+1)^2} = 1$

$\implies -6k=9$
$\implies k = -\frac{3}{2}$
Title: Re: Differentiation help please?
Post by: username on June 19, 2008, 06:25:18 pm: Thank you. ^_^