Question

Produce graphs of f that reveal all the important aspects of the curve. Then use calculus to find the following. (Enter your answers using interval notation. Round your answers to two decimal places.)
f(x) = 6 sin x + cot x, −π ≤ x ≤ π
Find the interval of increase.

Find the interval of decrease.

Find the inflection points of the function.

Find the interval where the function is concave up.



Find the interval where the function is concave down

Answer 1

\({\Huge\bbox[1pt,#ccffff ,border:2pt solid black ]{ _{_{_{\rm In\quad general:}}} }}\)

Suppose there is a differentiable function,
in a form of\(\rm : \) \(\large\color{black}{ \displaystyle y=f(x) }\)

You know that when the slope of \(\color{black}{f}\) is positive
the function is going up (or, "increasing").
\(\color{black}{\bf and}\) When the slope of \(\color{black}{f}\) is negative the
function is going down (or, "decreasing").
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You have probably heard that the derivative
of the function is the slope of the function.

\(\color{black}{\bf MORE\quad PRECISELY:}\)
For any function \(\color{black}{f(x)}\), there is a derivative
(which is also a function of x) that is denoted
as \(\color{black}{f'(x)}\) \([\)or denoted as \(\color{black}{y'}\) or otherwise\(]\) and
this "derivative" \(\color{black}{f'(x)}\) generates slope at any
point on the function \(\color{black}{f(x)}\).

\(\color{black}{\bf Consequentially:}\)
\(\color{blue}{[1]}\) When \(\color{black}{f'(x)>0}\) the slope is positive
- the \(\color{black}{f(x)}\) is increasing.

\(\color{blue}{[2]}\) When \(\color{black}{f'(x)<0}\) the slope is negative
- the \(\color{black}{f(x)}\) is decreasing.
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Now, another thing would be, to figure out
the "concavity".

\(\color{blue}{[1]}\) When the function's slope, or the
derivative \(\color{black}{f'(x)}\) increases, THEN the \(\color{black}{f(x)}\)
is said to be \(\color{black}{{\tt concave\color{white}{x}up}}\).

\(\color{blue}{[2]}\) When the function's slope, or the
derivative \(\color{black}{f'(x)}\) decreases, THEN the \(\color{black}{f(x)}\)
is said to be \(\color{black}{{\tt concave\color{white}{x}down}}\).
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\(\large\color{#e60000}{{\cal Closer\color{white}{x}Attention\color{white}{x}Please...}}\)

\(\color{red}{{\bf \left(Statement~1\right)}}\)
The ((first)) derivative \(\color{black}{f'(x)}\) can be used to
determine if \(\color{black}{f(x)}\) increases/decreases.

\(\color{red}{{\bf \left(Statement~2\right)}}\)
The ((second)) derivative \(\color{black}{f''(x)}\) can be used
to determine if \(\color{black}{f'(x)}\) increases or decreases.

\(\color{black}{{\tt Think\color{white}{x}about\color{white}{x}it,}}\) aren't these 2 statements
equivalent in a sense? Do you agree with me?
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Well, if I was clear enough then you should
be capable of determining concave-up and
concave down intervals.
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Lastly, \(\color{black}{{\tt inflection\color{white}{x}points}}\)\(\rm : \)

The inflection point is a point where the \(\color{black}{f(x)}\)
changes concavity \([\)that is, When the function
changes from concave-down to concave-up
or vice versa\(]\).

It is an equivalent of saying that the inflection
point(s) of \(\color{black}{f(x)}\) occur(s) when \(\color{black}{f'(x)}\) changes
from increasing to decreasing (or vice versa).

For a point \(\color{black}{{\tt x=a}}\) to be an inflection point,

\(\color{blue}{[1]}\) \(\color{black}{f''(a)=0}\)
((The second derivative is zero at x=a))

\(\color{blue}{[2]}\) Concavity actually changes.
You have to verify that this point is not
just a local max/min (because then it is
certainly not the inflection point).

You can conduct a simple test for this.
Plug in values of x that are near b. For
example if you find that:

\(\color{black}{f''(a+c)>0}\) and,
\(\color{black}{f''(a-c)<0}\)

OR, vice versa:

\(\color{black}{f''(a+c)<0}\) and,
\(\color{black}{f''(a-c)>0}\)

(Where \(\color{black}{c}\) is an effeciently small constant,
lets for convenience define it to be positive)

THEN, the function indeed has an inflection
at \(\color{black}{{\tt x=a}}\).

\(\color{red}{\bf BUT}\)
if you find that \(\color{black}{f''(a+c)}\) and \(\color{black}{f''(a-c)}\)
are BOTH positive or BOTH negative, then
\(\color{black}{{\tt x=a}}\) is NOT an inflection point.
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\(\color{#996600}{\bf I~~hope~~this~~helps~~you~~out,~~good~~luck!}\)