Find the following for function #f(x)=\ln(x^4+1)#...?
- intervals of increase/decrease
- local min/max values
- intervals of concavity and inflection points
- intervals of increase/decrease
- local min/max values
- intervals of concavity and inflection points
2 Answers
For
Explanation:
The graph below vividly answers all your questions. Yet, there is a
formal way of finding all.
So, for
and at
Also, by successive differentiation, it is seen that
also 0 at x = 0. But
derivative
At
inflexion.
So,
and for
Of course, at x = 0, it is flat. At
nor convex.
The tangent y = 0 does not cross the the curve here, at x = 0 but it
does at the points of inflexion
this edition .are attributed to the nice observation in the comment by
Jim I hope that the answer is now alright, in all aspects.
graph{y-ln(1+x^4)=0 [-10, 10, -5, 5]}
See below.
Explanation:
The denominator is always positive, so the sign of
So,
# = (12x^6+12x^2-16x^6)/(x^4+1)^2#
# = (-4x^6+12x^2)/(x^4+1)^2#
# = (-4x^2(x^4+3))/(x^4+1)^2#
On
On
On
On
The inflection changes at