MATHS PROBLEM Terlampir.... Question from @46jekkoi

46jekkoi @46jekkoi

November 2019 1 86 Report

MATHS PROBLEM
Terlampir

MRizky099

Misal: $\displaystyle f(x) = \sqrt{x^2+4}+\sqrt{x^2-24x+153}$

$\displaystyle \min\{f(x)\} = \dots?$

Penyelesaian:

Mencari turunan $f(x)$

Turunan $f(x)$ bentuk $f(x) = \sqrt{u}$ adalah

$\displaystyle \boxed{f'(x) = \frac{u'}{2\sqrt{u}}}$

sehingga

$\displaystyle f(x) = \sqrt{x^2+4}+\sqrt{x^2-24x+153} \\ f'(x) = \frac{2x}{2\sqrt{x^2+4}}+\frac{2x-24}{2\sqrt{x^2-24x+153}} \\ f'(x) = \frac{x}{\sqrt{x^2+4}}+\frac{x-12}{2\sqrt{x^2-24x+153}} \\ f'(x) = \frac{x\sqrt{x^2-24x+153}+(x-12)\sqrt{x^2+4}}{\sqrt{\left(x^2+4\right)\left(x^2-24x+153\right)}}$

Cari titik stasioner $f(x) \to f'(x) = 0$

$\displaystyle f'(x) = 0 \\ \frac{x\sqrt{x^2-24x+153}+(x-12)\sqrt{x^2+4}}{\sqrt{\left(x^2+4\right)\left(x^2-24x+153\right)}} = 0$

Abaikan pembilang karena pembilang ≠ 0

$\displaystyle x\sqrt{x^2-24x+153}+(x-12)\sqrt{x^2+4} = 0$

untuk mempersingkat, dimisalkan

$\displaystyle a=x^2-24x+153 \\ b=x^2+4$

sehingga

$\displaystyle \left(x\sqrt{a}+(x-12)\sqrt{b}\right)^2 = 0 \\ ax^2+2x(x-12)\sqrt{ab}+b(x-12)^2 = 0 \\ \left(ax^2+b(x-12)^2\right)^2 = \left(-2x(x-12)\sqrt{ab}\right)^2 \\ a^2x^4+2abx^2(x-12)^2+b^2(x-12)^4 = 4abx^2(x-12)^2 \\ a^2x^4-2abx^2(x-12)^2+b^2(x-12)^4 = 0 \\ \Big(ax^2-b(x-12)^2\Big)^2 = 0 \\ ax^2-b(x-12)^2 = 0 \\ ax^2-b(x^2-24x+144) = 0 \\ (a-b)x^2+24bx-144b = 0 \\ \Big(x^2-24x+153-x^2-4\Big)x^2+24x(x^2+4)-144(x^2+4) = 0 \\ (-24x+149)x^2+24x^3+96x-144x^2-576 = 0 \\ -24x^3+24x^3+149x^2-144x^2+96x-576 = 0 \\ 5x^2+96x-576 = 0 \\ (x+24)(5x-24) = 0 \\ \begin{array}{lcl}x+24=0&\text{atau}&5x-24=0 \\ x=-24&\text{atau}&x=\frac{24}{5} \\ \bold{(TM)}&{}&{} \end{array}$

Uji $f'(x)$ dan abaikan pembilang karena pembilang pasti selalu positif (syarat fungsi bentuk akar)

$\displaystyle \begin{aligned} \{x<\frac{24}{5}\}&: x=0 \to (0)\sqrt{(0)^2-24(0)+153}+((0)-12)\sqrt{(0)^2+4} &= 0+(-) < 0 \\ \{x>\frac{24}{5}\}&: x=12 \to (12)\sqrt{(12)^2-24(12)+153}+((12)-12)\sqrt{(12)^2+4} &= (+)+0 > 0 \end{aligned}$

Dari uji titik $f'(x)$ , ketika diilustrasikan akan seperti ini dalam bentuk garis bilangan:

$\displaystyle \boxed{\:\:\:\text{turun (-)}\:\:\:}\frac{24}{5}\boxed{\:\:\:\text{naik (+)}\:\:\:}$

Dilihat dari garis bilangan $f'(x)$ , nilai $\min\{f(x)\}$ didapat ketika $x=\frac{24}{5}$ sehingga nilai $\min\{f(x)\}$

$\displaystyle \beginno numeric noise key 1053\min\no numeric noise key 1052 &= f\left(\fracNO NUMERIC NOISE KEY 1051no numeric noise key 1050\right) \\ \sqrt{\left(\fracno numeric noise key 1049no numeric noise key 1048\right)^2+4}+\sqrt{\left(\fracno numeric noise key 1047no numeric noise key 1046\right)^2-24\left(\fracno numeric noise key 1045no numeric noise key 1044\right)+153} \sqrt{\fracno numeric noise key 1043no numeric noise key 1042}+\sqrt{\fracno numeric noise key 1041no numeric noise key 1040\left(\fracno numeric noise key 1039no numeric noise key 1038\right)+153} 1037no numeric noise key 1036}+\sqrt{\frac{-2304+3825}no numeric noise key 1035} \fracno numeric noise key 1034no numeric noise key 1033+\sqrt{\fracno numeric noise key 1032no numeric noise key 1031} 1030no numeric noise key 1029+\fracno numeric noise key 1028no numeric noise key 1027 1026no numeric noise key 1025 13 \endno numeric noise key 1024\\$