please help not sure abt the anwers..
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.
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.
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?
(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
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
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.