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4.5-bo'lim : Grafik shakli, I qism

10. \(h\left( t \right) = {t^2}\,\sqrt[3]{{t - 7}}\) 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 hosila shu erda.

\[\begin{align*}h'\left( t \right) & = 2t{\left( {t - 7} \right)^{{\frac{1}{3}}}} + {t^2}\left( {{\frac{1}{3}}} \right){\left( {t - 7} \right)^{ - \,\,{\frac{2}{3}}}} = 2t{\left( {t - 7} \right)^{{\frac{1}{3}}}} + \frac{{{t^2}}}{{3{{\left( {t - 7} \right)}^{{\frac{2}{3}}}}}}\\ & = \frac{{6t\left( {t - 7} \right) + {t^2}}}{{3{{\left( {t - 7} \right)}^{{\frac{2}{3}}}}}} = \frac{{7{t^2} - 42t}}{{3{{\left( {t - 7} \right)}^{{\frac{2}{3}}}}}} = \frac{{7t\left( {t - 6} \right)}}{{3{{\left( {t - 7} \right)}^{{\frac{2}{3}}}}}}\end{align*}\]

Endi eslang, tanqidiy nuqtalar lotin mavjud bo'lmagan yoki nolga teng bo'lgan joydir. Biz hosilani iloji boricha soddalashtirganimiz va faktorlarga ajratganimiz uchun hosila \(t = 7\) da mavjud emasligini (va funksiya bu yerda mavjud...) va \(t = 0\) va \(t = 6\) da hosila nolga teng ekanligini aniq ko‘ramiz. Bu funktsiyaning kritik nuqtalari quyidagilardan iborat:

\[\require{bbox} \bbox[2pt,border:1px solid black]{{t = 0,\,\,\,\,\,\,t = 6,\,\,\,\,\,\,t = 7}}\]

b Funksiyaning ortishi va kamayishi oraliqlarini aniqlang. Yechim

Funktsiyani oshirish/kamaytirish 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 \(h\left( t \right) = {t^2}\sqrt[3]{{t - 7}}\) with dashed vertical lines at \(t = 0\), \(t = 6\) and \(t = 7\) dividing it into four regions. In the first region \(h'\left( { - 1} \right) = 4.08\) so \(h'\left( t \right) > 0\), in the second \(h'\left( 1 \right) = - 3.53\) so \(h'\left( t \right) < 0\), in the third, whose computation is boxed and joined to its region by an arrow, \(h'\left( {6.5} \right) = 12.04\) so \(h'\left( t \right) > 0\), and in the fourth \(h'\left( 8 \right) = 37.33\) so \(h'\left( t \right) > 0\).

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

\[\require{bbox} \bbox[2pt,border:1px solid black]{{{\mbox{Increasing : }}\left( { - \infty ,0} \right),\,\,\,\,\,\,\left( {6,7} \right)\,\,\,\,\& \,\,\,\left( {7,\infty } \right)\hspace{0.25in}\hspace{0.25in}{\mbox{Decreasing : }}\,\,\left( {0,6} \right)}}\]

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.

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}t & = 0\,\,\,\,\,\,:\,{\mbox{ Relative Maximum}}\\ t & = 6\,\,\,\,\,\,:\,{\mbox{ Relative Minimum}}\\ t & = 7\,\,\,\,\,\,:\,{\mbox{ Neither}}\end{align*}}\]