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zad.1
\(f(x)=\frac{1-2x+x^2}{2x}\\
f''(-1)=?\)

zad.2
\(f(x)=\frac{x^2-2x+1}{x^2-4}\) znajdz ekstremum

Proszę o pomoc.

patrycja12100 pisze: ↑31 maja 2023, 20:49 zad.1
\(f(x)=\frac{1-2x+x^2}{2x}\\
f''(-1)=?\)

\(f(x)=\frac{(1-x)^2}{2x}\\
f'(x)=\frac{2(1-x)\cdot (-1)\cdot 2x-2(1-x)^2}{4x^2}\\
f'(x)=\frac{-4x(1-x)-2(1-x)^2}{4x^2}\\
f'(x)=\frac{-2(1-x)(2x+1-x)}{4x^2}\\
f'(x)=\frac{2(x-1)(x+1)}{4x^2}\\
f'(x)=\frac{x^2-1}{2x^2}\\
f''(x)=\frac{2x\cdot 2x^2-4x(x^2-1)}{4x^4}\\
f''(-1)=\frac{-2\cdot 2}{4}=-1
\)

patrycja12100 pisze: ↑31 maja 2023, 20:49

zad.2
\(f(x)=\frac{x^2-2x+1}{x^2-4}\) znajdz ekstremum

\(x\neq 2\;\;\; \wedge \;\;x\neq -2\)
\(f'(x)=\frac{(2x-2)(x^2-4)-(x^2-2x+1)2x}{(x^2-4)^2}\\
f'(x)=\frac{2(x-1)(x^2-4)-2x(x-1)^2}{(x^2-4)^2}\\
f'(x)=\frac{2(x-1)(x^2-4-x(x-1)}{(x^2-4)^2}\\
f'(x)=\frac{2(x-1)(-4+x)}{x^2-4)^2}\\
f'(x)=0\iff x\in\{1,4\}\\
f'(x)>0\iff x\in (-\infty, 1)\cup (4,\infty)\setminus\{-2\}\\
f'(x)<0\iff x\in (1,4)\setminus\{2\}\\
f_{max}=f(1)=0\\
f_{min}=f(4)=0,75
\)

eresh pisze: ↑31 maja 2023, 21:21

patrycja12100 pisze: ↑31 maja 2023, 20:49

zad.2
\(f(x)=\frac{x^2-2x+1}{x^2-4}\) znajdz ekstremum

\(x\neq 2\;\;\; \wedge \;\;x\neq -2\)
\(f'(x)=\frac{(2x-2)(x^2-4)-(x^2-2x+1)2x}{(x^2-4)^2}\\
f'(x)=\frac{2(x-1)(x^2-4)-2x(x-1)^2}{(x^2-4)^2}\\
f'(x)=\frac{2(x-1)(x^2-4-x(x-1)}{(x^2-4)^2}\\
f'(x)=\frac{2(x-1)(-4+x)}{x^2-4)^2}\\
f'(x)=0\iff x\in\{1,4\}\\
f'(x)>0\iff x\in (-\infty, 1)\cup (4,\infty)\setminus\{-2\}\\
f'(x)<0\iff x\in (1,4)\setminus\{2\}\\
f_{max}=f(1)=0\\
f_{min}=f(4)=0,75
\)

Dziękuję za pomoc.