Firstly, the cube root of 17 is 2.571281591 which is 2.57 to 3.s.f.
Initially, I thought about just approaching this problem using the Newton-Raphson Method when x0=2. In which case; x^3=17
x^3-17=0
Using the Newton-Raphson iterative formula xn+1=xr-f(xn)/f’(xn)
f(x)=x^3-17
f’(x)=3x^2...
Thank you for your reply. Yes, I found found that with nonlinear inequalities. Graphs certainly have been helpful to my understanding this topic better. Although, I still find it a little confusing.
Thank you for your suggestion for the alternative method of solving the second part to this...
Thank you for your reply. For 1, yes sorry my workings should simplify -3x+2-1/x<0 by converting to a fractional form;
-3xx/x+2x/x-1/x<0
-3x^2+2x-1/x<0
Multiply both sides by -1 to reverse the inequality;
3x^2-2x+1/x>0
Combine the intervals x>0 and 0<x<2/3, the solution is 0<x<2/3.
Yes, I...
1. I think the question is asking where is the graph of |3x-2| below the graph of 1/x.
To sketch the graph of y= |3x-2| draw the line of y=3x-2 and reflect the section with negative y-values in the x-axis. Alternatively, I could set 3x-2 ≥0, meaning |3x-2|=3x-2 so draw the line of y=3x-2. Then...
cot^2θ+5cosecθ=4
cot^2θ+5cosecθ-4=0
cosec^2θ+5cosec-4-1=0
cosec^2θ+5cosec-5=0
Let u=cosecθ
u^2+5u-5=0
Solve using the quadratic formula;
u=(-5± 3√5)/2
u=(-5+ 3√5)/2=0.8541...
Substitute cosecθ=u
Therefore, cosecθ=0.8541
1/sinθ=0.8541
sinθ=1/0.8541=1.170... which is not true since sin x cannot be...
Thank you very much for your informative reply. Right, I may add the domain and range of each function for f(x) and g(x) in part a. Really, I used the notation fg(x) and gf(x) because this was what the question specified and additionally what I had been taught in my textbook, which explained...
1. a.
fg(x)=2(1/2(x-1))+1
fg(x)=2(x/2-1/2)+1
fg(x)=x-1+1
fg(x)=x
gf(x)=1/2((2x+1)-1)
gf(x)=1/2(2x+1-1)
gf(x)=x+1/2-1/2
gf(x)=x
The functions functions f(x) and g(x) are inverses of each other. This can be demonstarted by
f(x)=2x+1
y=2x+1
x=2y+1
x-1=2y
(x-1)/2=y
Thus, y=1/2(x-1) = g(x)
And...
Thank you for your reply, sorry I should have used brackets.
Question b:
y=(4x+17)/(x+2)
y=(4x+17)/(x+2)undergoes this first transformation to become y=(4x+17)/(x+2-4)
Transform -4 into a fraction, -4(x+2)/(x+2)=(-4x-8)/(x+2)
y=(4x+17)/(x+2)(-4x-8)/(x+2)
Since the denominators are equal combine...
a. y=x^2 undergoes transformation 1 to become y=(x+2)^2
y=x^2+2 undergoes transformation 2 to become y=3(x+2)^2
y=3(x+2)^2 undergoes transformation 3 to become y=3(x+2)^2+4
So would the equation of the resulting curve be y=3(x+2)^2+4? I am very uncertain when it comes to performing...