3.1 Derivaatan ja derivoituvuuden tarkastelua

301

a) 1

b) 0,25

c) -2

d) 0

302
I, IV

303

$f'\left(x\right)=4x-1$

$f'\left(3\right)=4\cdot3-1=11$

$f'\left(x\right)=\lim_{h\rightarrow0}\left(\frac{f\left(3+h\right)-f\left(3\right)}{h}\right)=\lim_{h\rightarrow0}\left(\frac{2\cdot\left(3+h\right)^2-\left(3+h\right)-15}{h}\right)=\lim_{h\rightarrow0}\left(\frac{2\cdot\left(9+5h+h^2\right)-3+h-15}{h}\right)=\lim_{h\rightarrow0}\left(\frac{2h^2+11h}{h}\right)=\lim_{h\rightarrow0}\left(2h+11\right)=2\cdot0+11=11$

304

I Ei

II Ei

III Kyllä

IV Ei

305

$5h^2-30h+65$

$f'\left(1\right)=65$

$f'\left(t\right)=15t^2-90t+140$

$f'\left(1\right)=15-90+140=65$

306

$f\left(x\right)=-\left(x-3\right)^2$

$f'\left(x\right)=-x^2+6x-9=-2x+6$
$f'\left(-2\right)=-2\cdot\left(-2\right)+6=10$

307

$f'\left(x\right)=-2x$

$f'\left(x\right)=-2\left(-2\right)=4$

$f'\left(x\right)=\lim_{h\rightarrow0}\left(\frac{f\left(-2+h\right)-f\left(-2\right)}{h}\right)=\lim_{h\rightarrow0}\left(\frac{\left(1-\left(-2+h\right)^2-\left(1-4\right)\right)}{h}\right)=\lim_{h\rightarrow0}\left(\frac{\left(1-\left(h^2-4h+4\right)\right)+3}{h}\right)=\lim_{h\rightarrow0}\left(\frac{4-\left(h^2-4h+4\right)}{h}\right)=\lim_{h\rightarrow0}\left(4-h-4+4\right)=4-0-4+4=4$

308

$f\left(x\right)=x^2+x-1$

$f'\left(x\right)=\lim_{h\rightarrow0^-}\left(\frac{f\left(1+h\right)-f\left(1\right)}{h}\right)=\lim_{h\rightarrow0^-}\left(\frac{\left(\left(1+h\right)^2+\left(1+h\right)-1\right)-1}{h}\right)=\lim_{h\rightarrow0^-}\left(\frac{\left(\left(h^2+2h+1\right)+\left(1+h\right)-1\right)-1}{h}\right)=\lim_{h\rightarrow0^-}\left(\frac{h^2+3h+2-1-1}{h}\right)=\lim_{h\rightarrow0^-}\left(h+3+2-1-1\right)=0+3+2-2=3$

$f\left(x\right)=-x^2+5x-3$

$\lim_{h\rightarrow0^+}\left(\frac{f\left(1+h\right)-f\left(1\right)}{h}\right)=\lim_{h\rightarrow0^+}\left(\frac{\left(-\left(1+h\right)^2+5\cdot\left(1+h\right)-3\right)-\left(-1+5-3\right)}{h}\right)=\lim_{h\rightarrow0^+}\left(\frac{\left(-\left(h^2+2h+1\right)+5+5h-3\right)-1}{h}\right)=\lim_{h\rightarrow0^+}\left(\frac{-h^2-2h+5h-1+5-3-1}{h}\right)=\lim_{h\rightarrow0^+}\left(\frac{-h^2+3h}{h}\right)=\lim_{h\rightarrow0^+}\left(-h+3\right)=-0+3=3$

$k=3$

$y_0=1$
$x_0=1$

$\left(y-y_0\right)=k\left(x-x_0\right)$

$y-1=3x-3$
$y=3x-2$

309

$f\left(x\right)=2x+1$

$\lim_{h\rightarrow0^-}^{ }\left(\frac{f\left(-1+h\right)-f\left(-1\right)}{h}\right)=\lim_{h\rightarrow0^-}^{ }\left(\frac{\left(2\cdot\left(-1+h\right)+1\right)+1}{h}\right)=\lim_{h\rightarrow0^-}^{ }\left(\frac{-2+2h+1+1}{h}\right)=\lim_{h\rightarrow0^-}^{ }\left(\frac{2h}{h}\right)=2$

$f\left(x\right)=x^2+3x+1$

$\lim_{h\rightarrow0^+}\left(\frac{f\left(-1+h\right)-f\left(-1\right)}{h}\right)=\lim_{h\rightarrow0^+}\left(\frac{\left(\left(-1+h\right)\right)^2+3\cdot\left(-1+h\right)+1-\left(-1-3+1\right)}{h}\right)=\lim_{h\rightarrow0^+}\left(\frac{\left(h^2-2h+1\right)+3h-3+1+1+3-1}{h}\right)=\lim_{h\rightarrow0^+}\left(\frac{h^2-2h+3h-2}{h}\right)=\lim_{h\rightarrow0^+}\left(\frac{h^2+h-2}{h}\right)=\lim_{h\rightarrow0^+}\left(h+1-2\right)=0+1-2=-1$
$-1\ne2$

Ei ole

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