# Some calculus questions (1 Viewer)

#### Ragerunner

A bit stuck here..

Find the equation of the tangent to x^3 + y^3 - xy^2 = 1 at (1,1)

Also, what is the easiest way to find out whether a function is continuous or differentiable when given a function equation?

Thanks!

#### :: ck ::

##### Actuarial Boy
mmm hint...

implicit differentiation... then plug 1,1 in....
---------

and to tell if a function is continuous at a given point, wot u need to do is use limits, say at the point where x = a

lim(x->a<sup>-</sup>)f(x) = f(a) = lim(x->a<sup>+</sup>)f(x)

in terms of differntiability at the given point (x=a again), it needs to be
1) continuous at x=a and
2) lim(x->a<sup>-</sup>)f'(x) = f'(a) = lim(x->a<sup>+</sup>)f'(x)

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#### wogboy

##### Terminator
Find the equation of the tangent to x^3 + y^3 - xy^2 = 1 at (1,1)
Implicit differentiation is the way to go, differentiate both sides with respect to x:

x^3 + y^3 - xy^2 = 1
3x^2 + 3y^2*dy/dx - (y^2 + dy/dx*2xy) = 0
dy/dx * (3y^2 - 2xy) = y^2 - 3x^2
dy/dx = (y^2 - 3x^2)/(3y^2 - 2xy)
subbing (x,y)=(1,1),
dy/dx = -2

so the gradient of the line is -2. Seeing as it goes through (1,1):
y = -2x + 3

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#### wogboy

##### Terminator
Also, what is the easiest way to find out whether a function is continuous or differentiable when given a function equation?
From the fundamental definition of the continuity:

A function f is continuous at x=a iff:

for every e>0 there exists a d>0 such that:

whenever |x - a| < d,
then |f(x) - f(a)| < e

So for example if you're asked to prove from the fundamental definition that y=x^2 is continuous at x=0:

whenever |x - 0| = |x| < d,
we require |x^2 - 0^2| = x^2 < e

since |x| >= x^2 for x near 0,
it will suffice to require that |x| < e, in which case we can choose:
d = e

therefore y=x^2 is continuous at x=0.
-----------------------------------------
Obviously this isn't the easiest way to work out if a function is continuous, you only use this method if you're asked to prove a function is continuous from the fundamental definition. Otherwise see if it is composed of functions you know that are continuous.

To prove a function f is differentiable at x=a, it is quite easier.

f is differentiable at x=a if:

lim (h->0) {f(a+h) - f(a)}/ h exists

#### McLake

##### The Perfect Nerd
Can you please post your uni questions in the extra-cirircular topic from now on, thanks ...

#### :: ck ::

##### Actuarial Boy
i thought we needed 2 know that as part of curve sketching in 4unit?

#### McLake

##### The Perfect Nerd
Originally posted by :: ryan.cck ::
i thought we needed 2 know that as part of curve sketching in 4unit?
Part i) Yes
Part ii) I don't think so ...

#### :: ck ::

##### Actuarial Boy
ah ok well we covered it in tutoring briefly so yeh.. =\

oh well doesn't hurt to know it for hsc #### McLake

##### The Perfect Nerd
Originally posted by :: ryan.cck ::
ah ok well we covered it in tutoring briefly so yeh.. =\

oh well doesn't hurt to know it for hsc And it may help at uni ...

#### Ragerunner

Thanks for the replies. Understand it now I got few more questions a bit confused with.

1) Determine the limit as x -> infinity of : sqroot(4x^2+7x-2) - (2x+1)

2) Determine the limit as x -> 0 of : (x^2)*sin(1/x)

3) f(x) = (x^3 - 6x^2 + 11x - 6) / (x-a)

This is not continuous at x = a. For which values of a is the discontinuity removable?

#### wogboy

##### Terminator
1) Determine the limit as x -> infinity of : sqroot(4x^2+7x-2) - (2x+1)
If you see something like this in the form of lim {x -> ?} (a - b), where both a and b go to infinity, the following is often a handy trick to use:

lim {x -> ?} a - b
= lim {x -> ?} (a^2 - b^2)/(a+b)

so
lim {x -> infinity} sqrt(4x^2 + 7x - 2) - (2x+1)
= lim {x -> infinity} (4x^2 + 7x - 2 - (2x + 1)^2)/(sqrt(4x^2 + 7x - 2) + 2x + 1)
= ...
= 3/4

2) Determine the limit as x -> 0 of : (x^2)*sin(1/x)
since -1 <= sin(1/x) <= 1 for all real x,
-x^2 <= x^2 * sin(1/x) <= x^2 for all real x,
so lim{x->0} -x^2<= lim{x->0} x^2 * sin(1/x) <= lim{x->0} x^2
0 <= lim{x->0} x^2 * sin(1/x) <= 0
so by the pinching theorem,
lim{x->0} x^2 * sin(1/x) = 0

f(x) = (x^3 - 6x^2 + 11x - 6) / (x-a)
In order for there to be a removable discontinuity at x=a, (x-a) must be a factor of the polynomial x^3 - 6x^2 + 11x - 6 (otherwise the graph will asymptote to +- infinity)

so you need to solve the polynomial
a^3 - 6a^2 + 11a - 6 = 0, using techniques from 3U HSC maths.