Asosiy tarkibga o'tish

4.5-bo'lim : Grafik shakli, I qism

7. \(y = 2{x^3} - 10{x^2} + 12x - 12\) uchun quyidagi savollarning har biriga javob bering.

  1. Funktsiyaning kritik nuqtalarini aniqlang.
  2. Funksiyaning ortishi va kamayishi oraliqlarini aniqlang.
  3. Kritik nuqtalarni nisbiy maksimal, nisbiy minimum yoki hech biri sifatida tasniflang.
a Funktsiyaning kritik nuqtalarini aniqlang. Yechim

Bizga 1 kerakst kritik nuqtalarni olish uchun lotin, shuning uchun bu erda.

\[\frac{{dy}}{{dx}} = 6{x^2} - 20x + 12\]

Endi eslang, tanqidiy nuqtalar lotin mavjud bo'lmagan yoki nolga teng bo'lgan joydir. Shubhasiz, bu hosila hamma joyda mavjud (bu ko'phad ....) va hosila bu holda faktorlarga ajratilmaganligi sababli, hosila nolga teng bo'lgan joyni topish uchun tez kvadrat formulasini bajarishimiz kerak bo'ladi. Funktsiyaning kritik nuqtalari:

\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = \frac{{5 \pm \sqrt 7 }}{3} = 0.78475,\,\,2.54858}}\]

b Funksiyaning ortishi va kamayishi oraliqlarini aniqlang. Yechim

Funktsiyaning o'sishi/kamayishi ma'lumotlarini aniqlash uchun bizga kerak bo'lgan narsa - hosila uchun tezkor raqam chizig'i. Mana raqam chizig'i.

A number line for the derivative of \(y = 2{x^3} - 10{x^2} + 12x - 12\) with dashed vertical lines at \(x = \frac{{5 - \sqrt 7 }}{3}\) and \(x = \frac{{5 + \sqrt 7 }}{3}\) dividing it into three regions. In the left region \({\left. {\frac{{dy}}{{dx}}} \right|_{x = 0}} = 12\) so \(\frac{{dy}}{{dx}} > 0\), in the middle region \({\left. {\frac{{dy}}{{dx}}} \right|_{x = 1}} = - 2\) so \(\frac{{dy}}{{dx}} < 0\), and in the right region \({\left. {\frac{{dy}}{{dx}}} \right|_{x = 3}} = 6\) so \(\frac{{dy}}{{dx}} > 0\).

Bundan biz funktsiya uchun quyidagi oshirish/kamayish ma'lumotlarini olamiz.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{{\mbox{Increasing : }}\left( - \infty ,\frac{5 - \sqrt 7}{3} \right)\,\,\,\,\& \,\,\,\left( \frac{5 + \sqrt 7}{3} ,\infty \right)\hspace{0.25in}\hspace{0.25in}{\mbox{Decreasing : }}\,\,\left( \frac{5 - \sqrt 7}{3},\frac{5 + \sqrt 7}{3} \right)}}\]

c Kritik nuqtalarni nisbiy maksimal, nisbiy minimum yoki hech biri sifatida tasniflang. Yechim

Oldingi bosqichdagi ma'lumotni oshirish/kamaytirish bilan biz 1-dan foydalanib, muhim nuqtalarni osongina tasniflashimiz mumkinst hosilaviy test. Bu erda kritik nuqtalar funktsiyalarining tasnifi.

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}x & = \frac{5 - \sqrt 7}{3} = 0.78475\,\,\,\,\,\,:\,{\mbox{ Relative Maximum}}\\ x & = \frac{5 + \sqrt 7}{3} = 2.54858\,\,\,\,\,\,:\,{\mbox{ Relative Minimum}}\end{align*}}\]