• anonymous
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
  • Stacey Warren - Expert
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  • chestercat
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  • SolomonZelman
\({\Huge\bbox[1pt,#ccffff ,border:2pt solid black ]{ _{_{_{\rm In\quad general:}}} }}\) 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"). ───────────────────── 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. ───────────────────── 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 \(\color{black}{{\tt concave\color{white}{x}down}}\). ───────────────────── \(\large\color{#e60000}{{\cal Closer\color{white}{x}Attention\color{white}{x}Please...}}\) \(\color{red}{{\bf \left(Statement~1\right)}}\) 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? ───────────────────── Well, if I was clear enough then you should be capable of determining concave-up and concave down intervals. ───────────────────── 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. ───────────────────── \(\color{#996600}{\bf I~~hope~~this~~helps~~you~~out,~~good~~luck!}\)

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