Asosiy tarkibga o'tish

3.4-bo'lim : Mahsulot va miqdor qoidasi

10. \(\displaystyle f\left( x \right) = \frac{{x - {x^2}}}{{1 + 8{x^2}}}\) qayerda ortib, kamayib borayotganini aniqlang.

Yechim

Bizga birinchi navbatda hosila kerak bo'ladi, bu bo'lim qoidasini talab qiladi, chunki hosila bizga funktsiyaning o'zgarish tezligini berishini bilamiz. Mana lotin.

\[\require{bbox} \bbox[2pt,border:1px solid black]{{f'\left( x \right) = \frac{{\left( {1 - 2x} \right)\left( {1 + 8{x^2}} \right) - \left( {x - {x^2}} \right)\left( {16x} \right)}}{{{{\left( {1 + 8{x^2}} \right)}^2}}} = \frac{{1 - 2x - 8{x^2}}}{{{{\left( {1 + 8{x^2}} \right)}^2}}}}}\]
2-qadam

Keyinchalik, funktsiya qayerda o'zgarmasligini bilishimiz kerak va shuning uchun biz qilishimiz kerak bo'lgan narsa lotinni nolga tenglashtirish va hal qilishdir. Bunday holda, har qanday haqiqiy son uchun maxraj hech qachon nolga teng bo'lmasligi aniq va shuning uchun surat nolga teng bo'lgan joyda hosila nolga teng bo'ladi. Demak, payni nolga tenglashtirib, yechish shuni beradi:

\[1 - 2x - 8{x^2} = - \left( {8{x^2} + 2x - 1} \right) = - \left( {4x - 1} \right)\left( {2x + 1} \right) = 0\]

Bundan hosila nolga teng bo'lishini va shuning uchun funktsiya o'zgarmasligini ko'rish juda oson,

\[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - \frac{1}{2} \hspace{0.25in} x = \frac{1}{4}}}\]
3-qadam

Ushbu muammoga javob olish uchun biz bilishimiz kerak bo'lgan narsa, hosila ijobiy (va shuning uchun funktsiya ortib bormoqda) yoki salbiy (va shuning uchun funktsiya kamayishi) qaerda ekanligini bilishimiz kerak. Hosil uzluksiz bo'lgani uchun biz bilamizki, uning belgisini o'zgartirishi mumkin bo'lgan yagona joy hosila nolga teng bo'lgan joydir. Shunday qilib, biz ushbu bo'limda qilganimizdek, tezkor raqamlar qatori bizga turli intervallar uchun hosilaning belgisini beradi.

Mana bu muammo uchun raqam chizig'i.

A number line for the derivative of \(f\left( x \right) = \frac{{x - {x^2}}}{{1 + 8{x^2}}}\) with dashed vertical lines at \(x = - \frac{1}{2}\) and \(x = \frac{1}{4}\) dividing it into three regions. In the left region \(f'\left( { - 1} \right) = - 0.06\) so \(f'\left( x \right) < 0\), in the middle region \(f'\left( 0 \right) = 1\) so \(f'\left( x \right) > 0\), and in the right region \(f'\left( 1 \right) = - 0.11\) so \(f'\left( x \right) < 0\).

Bundan biz quyidagi o'sish/kamayish ma'lumotlarini olamiz.

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}{\mbox{Increasing :}} & \,\, - \frac{1}{2} < x < \frac{1}{4}\\ {\mbox{Decreasing :}} & \,\, - \infty < x < - \frac{1}{2},\,\,\,\,\frac{1}{4} < x < \infty \end{align*}}\]