Matematik Opgaver 313-315 317 320


Indholdsfortegnelse

Opgave 313

LaTex: \begin{eqnarray} f_1 (x) &=& x\sqrt x \\ f_1 '(1) &=& 1 \cdot \sqrt 1 + 1 \cdot \frac{1}{{2\sqrt 1 }} = 1\frac{1}{2} \\ f_2 (x) &=& \sqrt x \times (3 - 2x) \\ f_2 '(1) &=& \frac{1}{{2\sqrt 1 }} \cdot (3 - 2 \times 1) + \left( { - 2 \cdot \sqrt 1 } \right) = - 1\frac{1}{2} \\ f_3 (x) &=& \left( {2x^2 - 5x} \right)\left( {x^2 + 6} \right) \\ f_3 '(1) &=& \left( {2 \times 2 \times 1 - 5 \times 1^2 + 6} \right) + \left( {2 \times 1 \times 2 \times 1^2 - 5 \times 1} \right) = 6 \\ f_4 (x) &=& \frac{1}{x} \times \sqrt x \\ f_4 '(1) &=& \frac{{ - 1}}{{1^2 }} \times \sqrt 1 + \frac{1}{1} \times \frac{1}{{2\sqrt 1 }} = - \frac{1}{2} \\ \end{eqnarray}

Opgave 314

LaTex: \begin{eqnarray} p(x) &=& a\cdot x^2+b\cdot x +c\\ p'(x) &=& 2\cdot a\cdot x+b\\ p'(x) &=& 0 = 2\cdot a\cdot x+b\\ x &=& \frac{b}{-2\cdot a}\\ \end{eqnarray}

Opgave 315

LaTex: \begin{eqnarray} f(x) &=& (x-1)^2+2\\ &=& x^2-2x+3\\ y(x) &=& \frac{12\cdot x-13}{4}\\ &=& \frac{12\cdot x}{4}-\frac{13}{4}\\ &=& 3\cdot x-\frac{13}{4}\\ 3\cdot x-\frac{13}{4}&=& x^2-2x+3\\ 0 &=& x^2-5\cdot x+6,25\\ x &=& solve(x^2-5\cdot x+6,25,x) = \frac{5}{2}\\ \end{eqnarray}

Opgave 317

LaTex: \begin{eqnarray} p(t) &=& 10^6 + 10^4 t - 10^3 t^2 \\ p'(t) &=& - 2 \times 10^3 t + 10^4 \\ p'(0) &=& - 2 \times 10^3 \times 0 + 10^4 = 10^4 \\ p'(5) &=& - 2 \times 10^3 \times 5 + 10^4 = 0 \\ p'(10) &=& - 2 \times 10^3 \times 10 + 10^4 = - 10000 \\ \end{eqnarray}

Opgave 320

LaTex: \begin{eqnarray} f(a) &=& f'(a)\cdot a + b\\ b &=& f(a)-f'(a)\cdot a\\ y(x) &=& f'(a)\cdot x + b\\ &=& f'(a)\cdot x + f(a)-f'(a)\cdot a\\ 0 &=& f'(a)\cdot t_x + f(a)-f'(a)\cdot a\\ t_x &=& -\frac{f(a)-f'(a)\cdot a}{f'(a)}\\ t_y &=& f'(a)\cdot 0 + f(a)-f'(a)\cdot a\\ &=& f(a)-f'(a)\cdot a\\ t_a &=& \frac{1}{2}\cdot t_x\cdot t_y\\ &=& \frac{-(f(a)-f'(a)\cdot a)^2}{2\cdot f'(a)}\\ f(x) &=& \frac{3}{x}\\ f'(x) &=& \frac{-3}{x^2}\\ t_a &=& \frac{-36}{-6} = 6\\ \end{eqnarray}

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