# Math Help - analysis questions

1. ## analysis questions

2. Originally Posted by bobby77
a) A tangent line can be used as a linear approximation to a function near
a point, but it is not the only such approximation, so the answer is no -
a linear approximation to a function is not necessarily a tangent line.

b) At least one of the definitions of a point of inflection for a function f is
that it is a point at which the f'' changes sign so of necessity f''(x)=0 at
a point of inflection, but it is not a sufficient condition. As Glaysher points
out every point on f(x)=mx+c is a point where f''(x)=0, but they are not
points of inflection. So f''(a)=0, is not sufficient to guarantee that x=a is
a point of inflection of f.

RonL

3. Originally Posted by bobby77
c) f''<0, means that f' is decreasing, which is concave downwards, to be
concave upwards we would need f''>0, so no.

RonL

4. Indeed, the limits for both f and g do equal e, hence they are both true.

The number e is defined by those limits.

5. Originally Posted by CaptainBlack
a) A tangent line can be used as a linear approximation to a function near
a point, but it is not the only such approximation, so the answer is no -
a linear approximation to a function is not necessarily a tangent line.

b) At least one of the definitions of a point of inflection for a function f is
that it is a point at which f''(x)=0, then yes if f''(a)=0, then x=a is a point
of inflection of f.

RonL
b) Is no as it has to be a stationary point as well or else every point on a straight line would be a point of inflexion as:

f(x) = mx + c
f'(x) = m
f''(x) = 0

6. Originally Posted by bobby77
d) Yes

e) No, depends on the shape of the graph

eg a graph with a high positive gradient that suddenly levels off at the top can produce bad approximations

f) No, limit is 1

g) Yes

h) True only if x is in radians so statement not specific enough

7. Originally Posted by Glaysher
b) Is no as it has to be a stationary point as well or else every point on a straight line would be a point of inflexion as:

f(x) = mx + c
f'(x) = m
f''(x) = 0
I was being a bit relaxed with the definition, a point of inflection is
a point where the second derivative changes sign, so of necessity
f''=0, but f''=0 is not a sufficient condition.

So you are right my answer was wrong f''(a)=0 is not a sufficient
condition to guarantee that x=a is a point of inflection.

But a point of inflection does not have to be a stationary point.

RonL

8. Originally Posted by CaptainBlack
I was being a bit relaxed with the definition, a point of inflection is
a point where the second derivative changes sign, so of necessity
f''=0, but f''=0 is not a sufficient condition.

It does not have to be a stationary point.

RonL
Never seen a proper definition for it before

9. Originally Posted by CaptainBlack
In fact having just typed that definition it struck me that f'' can change
sign if it has a jump discontinuity at a point, is this still a point of inflection?

If so f'' !=0 at this point?

RonL
I suspect that the full formal definition would need the function to be continuous over some interval containing the point.

But just to confirm would there be a point of inflexion for a function that starts with a decreasing positive gradient which later turns into an increasing positive gradient before the gradient is zero?

10. Originally Posted by Glaysher
I suspect that the full formal definition would need the function to be continuous over some interval containing the point.

But just to confirm would there be a point of inflexion for a function that starts with a decreasing positive gradient which later turns into an increasing positive gradient before the gradient is zero?
Yes.

(I wouldn't pay too much attention to the post of mine that you quoted,
as I think a jump discontinuity in the second derivative implies a corner
in the derivative at the point, which is a point at which the second derivative
does not exist. My comment was an unjustified outbreak of mathematical
paranoia )

RonL

11. Originally Posted by CaptainBlack
Yes.

(I wouldn't pay too much attention to the post of mine that you quoted,
as I think a jump discontinuity in the second derivative implies a corner
in the derivative at the point, which is a point at which the second derivative
does not exist. My comment was an unjustified outbreak of mathematical
paranoia )

RonL
I was thinking that it would be unusual to have a point of inflexion on a point that was not part of the original graph

12. Originally Posted by Glaysher
I was thinking that it would be unusual to have a point of inflexion on a point that was not part of the original graph
You initial beleif that a point of inflection has to be a stationary point of
the function is quite interesting, as it is what I thought up to only a few
years ago. I suspect it is something left over as a half memory from
A-level maths 35+ years ago.

It just goes to show that one always need to be verifying what one
thinks one knows

RonL

13. Originally Posted by CaptainBlack
You initial beleif that a point of inflection has to be a stationary point of
the function is quite interesting, as it is what I thought up to only a few
years ago. I suspect it is something left over as a half memory from
A-level maths 35+ years ago.

It just goes to show that one always need to be verifying what one
thinks one knows

RonL
True as it was my A level teaching that made me think that too and the textbooks I currently teach from too. Though to be fair it is never explicitly said, just implied.

14. Originally Posted by Glaysher
d) Yes

e) No, depends on the shape of the graph

eg a graph with a high positive gradient that suddenly levels off at the top can produce bad approximations

f) No, limit is 1

g) Yes

h) True only if x is in radians so statement not specific enough
How is the limit in part f 1? Care to explain? I got e, although I might have done some careless mistake.

15. Originally Posted by AfterShock
How is the limit in part f 1? Care to explain? I got e, although I might have done some careless mistake.
If the problem is:
limit(x->0) (1+x)^(1/x) (it was hard to read the exponent, is this right?)

then the answer is indeed "e" not 1.

We can change the limit variable to n = 1/x, then the limit becomes:

limit(n->infinity) (1 + [1/n])^n which is one of the definitions of e.

-Dan

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