Also, do you know how to answer question 13 from 10D. Thank you.
Also, do you know how to answer question 13 from 10D. Thank you.
Sorry, the question is:
Find positive integers a and b such that
x^2y'' + 2xy' = 12y, where y = x^a + x^-b.
Thank you.
Thanks for the help with that question. Could you also explain Q 11 b)
Find the nth and (n+1)th derivative of x^n.
I understand that the nth derivative is n(n-1)(n-2) ... x^n-n = n(n-1)(n-2)...1
but am unsure why the (n+1)th derivative is 0. I assume it has something to do with x^n-(n+1). But that equals x^-1.
Thank you a lot for you help.
Cheers.
Thanks. Also for 12c)
Find the y' and y'' in terms of t:
c) x = 1 - 5t , y =t^3.
y' = -3/5 t^2
then isn't y'' = -6/5t
the answer in the book say y'' = 6/25t
Thanks.
i still don't understand. I get the same problem in question e)
x = (t -2)^2 , y = 3t
y' = 3/2 ( t -2)^-1
and i get
y'' = -3/2 (t-2)^-2
but the answer for y'' is -3/4 (t-2)^-3
Cheers.
Thank you for your help with that question.
For Q5) of 10E
Find the range of value of x for which the curve y = 2x^3 -3x^2 -12x + 8 is:
a) increasing, b)decreasing, c) concave up, d)concave down.
For c) and d), i know you find the y'' and then for c) make it > 0 and d) make it < 0. But am unsure about how to work out increasing and decreasing.
Thank you. Cheers.
For Q7)
Sketch possible graphs of continuous functions with these properties:
a) f(-5) = f(0) = f(5) = 0 , and f'(3) = f'(-3) = 0, and f''(x) > 0 for x < 0, and f''(x) <0 for x > 0
What I don't get is the first bit, is it saying that the x cord. = -5, 0, 5 has y cord = 0 ?
whereas the second bit states that at the point 3 and -3 there is a stationary point.
Cheers.
EDIT: I misinterpreted your post
Yes it does mean that the coordinates are (-5,0), (5,0) and (0,0) < --- The x-intercepts
There are stationary points at x = 3 and x = -3 but since you have been given insufficient information, you can plot these stationary points with any y-coordinate (But the x-coordinates have to stay the same as mentioned above) and it should be enough to get you the mark correct (Make it reasonable though by placing it within the scope of your axis - don't unnecessarily place the y-coordinate at like 1000) . Furthermore, you must be able to plot it with the correct concavity (i.e. Concave up or down).
Also
f''(x) > 0 means that it is a concave up for x < 0
f''(x) < 0 means that it is concave down for x > 0
^ This should give you an idea of the concavity of the stationary points - whether it is minimum or maximum
Therefore:
X-Intercepts:
x = -5
x = 0
x = 5
Minimum Turning Point: x = 3 (Any y-coordinate)
Maximum Turning Point: x = -3 (Any y-coordinate)
You should be able to draw it now, so good luck
Does anyone know how to answer:
Q12 from 10C:
a) Differentiate y = x^1/2 - x^3/2
b) find those values of x for which y' = 0, and hence determine the coordinates of any critical points.
c) Hence sketch a graph of y^2 = x(1-x)^2.
Cheers.
I am having difficulties doing question 5g from 10F
Without finding inflexions ,sketch the graphs of the function: Indiciate any asymptotes, stationary poitns and intercepts with the axes.
g) y = x^2 + 1/x^2
I don't know how to find the oblique parabola asymptote. Also for some reason I am getting one stationary point instead of the required two.
Cheers.
Put it under a common denominator
y = (x^4 + 1) / x^2
Like you would find the vertical asymptote, take the limit of x approaches infinity of the above equation
In these situations when there's a pronumeral in the denominator you should be expecting an oblique asymptote rather than a straight line
Then divide the numerator and denominator by the HIGHEST POWER OF X IN THE DENOMINATOR
lim
x --> Infinity [(x^4 / x^2) + (1 / x^2)] / (x^2 / x^2)
Simplify this
lim
x --> Infinity [(x^2) + (1 / x^2)] / (1)
Now if you use the limit you'll notice that as x approaches infinity the 1 / x^2 will equal to zero (Sub a huge number into x for this and you will notice it)
Therefore
y = [(x^2) + (0)] / (1)
y = x^2 / 1
y = x^2
Also don't forget the asymptote at x = 0 !!!
As for the stationary points
Stationary points when dy/dx = 0
y' = 2x - (2 / x^3)
0 = 2x - (2 / x^3)
Multiply both sides by x^3
0 = 2x^4 - 2
Factorise
0 = 2(x^4 - 1)
0 = 2(x^2 - 1)(x^2 + 1)
0 = 2(x - 1)(x + 1)(x^2 + 1)
Therefore
x = 1 or x = -1
Continue from here ~
Firstly, notice that the function f, defined as is positive for all . Also, note that f(-x) = f(x) which means that f is an even function, so the graph of f will be reflected across the y-axis. You could set the first derivative equal to 0, so that will give you the stationary points, namely at . Then sketch the graph accordingly. Also note that as x tends to infinity, your y values will start to tend to and as x tends to 0, your y values will tend to so these will essentially become your oblique asymptotes for the graph of f.
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