Prelim 2016 Maths Help Thread (3 Viewers)

kawaiipotato · Jun 26, 2016

eyeseeyou said:
Consider the function f(x)=(x^3+8x)/8
a. Show that f(x) is an increasing function
b. Find the equation of the tangent to y=f(x) at the origin
c. Using equal scales of axes, sketch a graph of y=f(x) over the domain -2=<x=<2 and draw its tangent at the origin
d. explain why an inverse exists and draw a grapg of y=f^-1(x) on your diagram
e. Solve f^-1(x)=8
f. Evaluate Integral of 3 to 0 f^-1(x) dx

$$a.$ \ $To assess if a function is "increasing" or "decreasing" then we need to assess the derivative.$

$f(x) = \frac{x^3 + 8x}{8}$

$f'(x) = \frac{1}{8}(3x^2 + 8)$

$The derivative is a positive value for all real x in its domain. What does this mean?$

$$b.$ \ f'(x) = \frac{1}{8}(3x^2 + 8)$

$f'(0) = 1$

$$When$ \ x = 0, \ f(0) = 0$

$Equation of the tangent at a point$: \ y - y_1 = m(x-x_1) \ \ \ $Where$ \ (x_1 , y_1 ) \ $is a point satisfying the function and m is the derivative of the function at that point$

$d.$ \ $An inverse of a function exists if the function is surjective (onto, but we can safely assume it is) and injective (one-to-one).$ \ $We can use the fact that the function is always increasing throughout its domain and so it will be one-to-one.$

$$e.$ \ $If we have an invertible function$ \ y = f(x) \ $and we want to solve$ f^{-1}(x) = a \ $where a is a constant$, \ $then this is just equal to$ \ f(a)$

$$So if$ \ f^{-1}(x) = 8 \ $then$ \ f(f^{-1}(x)) = f(8) \rightarrow x = f(8) = 72$

$$f.$ \ $Drawing up a graph, we can see that$ \ \int_{0}^{3} f^{-1}(x) dx = $Area of the rectangle enclosed by$ (0,0),(3,0),(3,f^{-1}(3)),(0,f^{-1}(3)) - \int_{0}^{f^{-1}(3)} f(x) \ dx$

$$For$ \ f^{-1}(3), \ $we let$ f(x) = 3 \ $and solve for$ \ x$

$3 = \frac{1}{8}(x^3 + 8x)$

$x^3 + 8x - 24 = 0$

$x = 2 \ $is the only real root$$

$$Hence$ \ f^{-1}(3) = 2$

$$So we just need to compute$ \ $Area of the rectangle enclosed by$ (0,0),(3,0),(3,2),(0,2) - \int_{0}^{2} \frac{1}{8}(x^3 + 8x) \ dx$

eyeseeyou said:
BTW Integrand just realised but you've been here all night lol

So?

Trebla · Jun 26, 2016

eyeseeyou said:
How do I differentiate surds?

The fact that you don't know this and want to 4U is a little concerning

Green Yoda · Jun 26, 2016

ICU trebla has asked you a question multiple times and very politely, and you have ignored. How do you expect us to answer about 60 question at once when you can't answer one?

DatAtarLyfe · Jun 26, 2016

eyeseeyou said:
Consider the function f(x)=(x^3+8x)/8
a. Show that f(x) is an increasing function
b. Find the equation of the tangent to y=f(x) at the origin
c. Using equal scales of axes, sketch a graph of y=f(x) over the domain -2=<x=<2 and draw its tangent at the origin
d. explain why an inverse exists and draw a grapg of y=f^-1(x) on your diagram
e. Solve f^-1(x)=8
f. Evaluate Integral of 3 to 0 f^-1(x) dx

What parts of this don't you get?
Becoz a-c is basic for prelim

DatAtarLyfe · Jun 26, 2016

DatAtarLyfe said:
What parts of this don't you get?
Becoz a-c is basic for prelim

Edit: nvm, just saw kawaiis post

leehuan · Jun 26, 2016

kawaiipotato said:
$d.$ \ $An inverse of a function exists if the function is surjective (onto, but we can safely assume it is) and injective (one-to-one).$ \ $We can use the fact that the function is always increasing throughout its domain and so it will be one-to-one.$

Small note - Because we don't define surjection in the HSC course, all we need to worry about is injection.

(Unless you forgot this is HSC and not MATH1151

)

Simorgh · Jun 26, 2016

I cant believe I made this thread and then hardly ever use it to help myself lel

Nailgun · Jun 26, 2016

Simorgh said:
I cant believe I made this thread and then hardly ever use it to help myself lel

Well not needing help isn't really a bad thing ahahah

fluffchuck · Jun 26, 2016

Hey guys, how would you tackle this question?

It was proven in the notes above that the tangents to the parabola x2 = 4ay at two points
P(2ap, ap^2) and Q(2aq, aq^2) on the parabola intersect at the point M(a(p + q), apq).
Explain why this result can be restated as follows: ‘The tangents at two points on the
parabola x^2 = 4ay meet at a point whose x-coordinate is the arithmetic mean of the
x-coordinates of the points, and whose y-coordinate is one of the geometric means of the
y-coordinates of the points.’ Which geometric mean is it?

Thank you!

InteGrand · Jun 26, 2016

fluffchuck said:
Hey guys, how would you tackle this question?

It was proven in the notes above that the tangents to the parabola x2 = 4ay at two points
P(2ap, ap^2) and Q(2aq, aq^2) on the parabola intersect at the point M(a(p + q), apq).
Explain why this result can be restated as follows: ‘The tangents at two points on the
parabola x^2 = 4ay meet at a point whose x-coordinate is the arithmetic mean of the
x-coordinates of the points, and whose y-coordinate is one of the geometric means of the
y-coordinates of the points.’ Which geometric mean is it?

Thank you!

$\noindent The geometric mean of two non-negative real numbers $a$ and $b$ is defined as $\sqrt{ab}$.$

fluffchuck · Jun 26, 2016

InteGrand said:
$\noindent The geometric mean of two non-negative real numbers $a$ and $b$ is defined as $\sqrt{ab}$.$

Woah that was really quick, thank you very much!

jathu123 · Jun 27, 2016

Not really a help post, but I found a couple of interesting trig questions if any of you prelim people wanna try it out

$$\noindent$ 1)$ $ $solve $ 5\sin ^{2}x-2\sin 2x +3\cos^{2}x=2 $ for $ 0\leq x\leq 360^{\circ}\\ \\ 2) $ show that $ \frac{1+\sin 2x}{1+\cos 2x}\equiv \frac{1}{2}(1+\tan x)^{2}$

leehuan · Jun 27, 2016

Hint for #2

jathu123 said:
Not really a help post, but I found a couple of interesting trig questions if any of you prelim people wanna try it out

$$\noindent$ 1)$ $ $solve $ 5\sin ^{2}x-2\sin 2x +3\cos^{2}x=2 $ for $ 0\leq x\leq 360^{\circ}\\ \\ 2) $ show that $ \frac{1+\sin 2x}{1+\cos 2x}\equiv \frac{1}{2}(1+\tan x)^{2}$

Choose the more convenient cosine double angle expansion.

P.S. what is 1 equal to?

Paradoxica · Jun 27, 2016

leehuan said:
Hint for #2

Choose the more convenient cosine double angle expansion.

P.S. what is 1 equal to?

stop spoiling it for others...

Green Yoda · Jun 27, 2016

1) x=90deg?

jathu123 · Jun 27, 2016

Rathin said:
1) x=90deg?

Unfortunately nope, have another try haha

Green Yoda · Jun 27, 2016

my bad x= 45deg or 225deg

jathu123 · Jun 27, 2016

Rathin said:
my bad x= 45deg or 225deg

Yup! There's also an extra 2 solution

eyeseeyou · Jul 2, 2016

The wording of this question is confusing the hell out of me:

If f(x)=e^(x+1) find the inverse function f^-1(x) and hence show that f(f^-1(x))=f^-1(f(x))=x

So far this was what I did:

f(x)=e^(x+1)
y=e^(x+1)
x=e^(y+1)
lnx=y+1
y=lnx-1
f^-1(x)=lnx-1

Thanks

leehuan · Jul 2, 2016

eyeseeyou said:
The wording of this question is confusing the hell out of me:

If f(x)=e^(x+1) find the inverse function f^-1(x) and hence show that f(f^-1(x))=f^-1(f(x))=x

So far this was what I did:

f(x)=e^(x+1)
y=e^(x+1)
x=e^(y+1)
lnx=y+1
y=lnx-1
f^-1(x)=lnx-1

Thanks

And you're still trying to do inverse functions before knowing what actual functions are.

$\text{We have }f(x)=e^{x+1}\\ \text{So }f(f^{-1}(x))=e^{f^{-1}(x)+1}=e^{(\ln x - 1)+1} = e^{\ln x} = x$

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