We found that \(f\) was decreasing on "both sides of \(x=1.\)"Find the intervals on which \(f(x) = x^{8/3}-4x^{2/3}\) is increasing and decreasing and identify the relative extrema.We again start with taking derivatives. A function is \"increasing\" when the y-value increases as the x-value increases, like this:It is easy to see that y=f(x) tends to go up as it goes along.
(c) The graph of h gets lower from left to right on part of its domain and gets higher from left to right on another part of its domain.
(a) Find the largest interval on which f is increasing. To find intervals on which \(f\) is increasing and decreasing:We demonstrate using this process in the following example.Let \(f(x) = x^3+x^2-x+1\). \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)[ "article:topic", "First Derivative Test", "decreasing on the interval I", "increasing on the interval I", "authorname:apex", "showtoc:no", "license:ccbync" ][ "article:topic", "First Derivative Test", "decreasing on the interval I", "increasing on the interval I", "authorname:apex", "showtoc:no", "license:ccbync" ]\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)THeorem \(\PageIndex{1}\): Test For Increasing/Decreasing FunctionsKey Idea 3: Finding Intervals on Which \(f\) is Increasing or DecreasingExample \(\PageIndex{1}\): Finding intervals of increasing/decreasingExample \(\PageIndex{2}\): Using the First Derivative TestExample \(\PageIndex{3}\): Using the First Derivative Test
But note:\[\dfrac{f(b)-f(a)}{b-a} \Rightarrow \dfrac{\text{numerator }>0}{\text{denominator } >0} \Rightarrow \text{slope of the secent line} >0 \Rightarrow \text{Average rate of chjange of $f$ on $[a,b]$ is $>0$.} This leads us to the following method for finding intervals on which a function is increasing or decreasing.Let \(f\) be a differentiable function on an interval I. We have \(f'(x) = 3x^2+2x-1 = (3x-1)(x+1)\), so \(f'(x) = 0\) when \(x=-1\) and when \(x=1/3\). Search.
This new understanding of increasing and decreasing creates a great method of determining whether a critical point corresponds to a maximum, minimum, or neither. Since we know we want to solve \(f'(x) = 0\), we will do some algebra after taking derivatives.\[\begin{align} f(x) &= x^{\frac{8}{3}}-4x^{\frac{2}{3}} \\ f'(x) &= \dfrac{8}{3} x^{\frac{5}{3}} - \dfrac{8}{3}x^{-\frac{1}{3}} \\ &= \dfrac{8}{3}x^{-\frac{1}{3}} \left(x^{\frac{6}{3}}-1 \right)\\ &=\frac{8}{3}x^{-\frac{1}{3}}(x^2-1)\\ &=\frac{8}{3}x^{-\frac{1}{3}}(x-1)(x+1).
Imagine a function increasing until a critical point at \(x=c\), after which it decreases. A quick sketch helps confirm that \(f(c)\) must … Key Idea 3 describes how to find intervals where \(f\) is increasing and decreasing Since \(f\) is not defined at \(x=1\), the increasing/decreasing nature of \(f\) could switch at this value. We give three reasons why the above work is worthwhile.First, the points at which \(f\) switches from increasing to decreasing are not precisely known given a graph. Also, Figure \(\PageIndex{4}\) shows a graph of \(f\), confirming our calculations. \(f'\) is never undefined.Since an interval was not specified for us to consider, we consider the entire domain of \(f\) which is \((-\infty,\infty)\). If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
(c) As can be seen from the graph above, [5, 6] is the largest interval containing 6 on which f is decreasing. This figure also shows \(f'\), again demonstrating that \(f\) is increasing when \(f'>0\) and decreasing when \(f'<0\).One is often tempted to think that functions always alternate "increasing, decreasing, increasing, decreasing,\(\ldots\)" around critical values. Theorem \(\PageIndex{1}\) below turns this around by stating "If \(f'\) is postive, then \(f\) is increasing."
At \(x=3\), the sign of f'\ switched from negative to positive, meaning \(f(3)\) is a relative minimum.
Give an example of functions f and g for which the statement is true, or say why such an ex…
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