Question 1163542
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*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f(x)\ =\ \frac{2}{5}x^5\,-\,\frac{1}{4}x^4\,-\,\frac{1}{3}x^3\ -\ 8]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{df}{dx}\ =\ 2x^4\ -\ x^3\ -\ x^2]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2f}{dx^2}\ =\ 8x^3\ -\ 3x^2\ -\ 2x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 2x^4\ -\ x^3\ -\ x^2\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x^2(2x\,+\,1)(x\,-\,1)\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{df}{dx}\|_{-\frac{1}{2}}\ =\ \frac{df}{dx}\|_{0}\ =\ \frac{df}{dx}\|_{1}\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2f}{dx^2}\|_{0}\ =\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2f}{dx^2}\|_{-\frac{1}{2}}\ =\ -\frac{3}{4}\ <\ 0]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d^2f}{dx^2}\|_{1}\ =\ 3\ >\ 0]


Intervals of increase:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(-\infty,-\frac{1}{2}\),\ \(1,\infty\)]


Interval of decrease


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \(-\frac{1}{2},1\)]


Local Maximum: 


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ -\frac{1}{2}\ \Left\ \frac{df}{dx}\,=\,0,\ \frac{d^2f}{dx^2}\,<\,0]


Local Minimum


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 1\ \Left\ \frac{df}{dx}\,=\,0,\ \frac{d^2f}{dx^2}\,>\,0]


Point of inflection


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ x\ =\ 0\ \Left\ \frac{df}{dx}\,=\,0,\ \frac{d^2f}{dx^2}\,=\,0]

								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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