Matematik Opgaver 326 333 335

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Opgave 326

1: $LaTex: \begin{eqnarray} f(x) &=& 5\cdot \sqrt{x}-2\cdot x^3+2\\ f'(x) &=& \frac{5}{2\cdot \sqrt{x}}-6\cdot x^2\\ \end{eqnarray}$

2: $LaTex: \begin{eqnarray} f(x) &=& \frac{\sqrt{x}+x^2}{2\cdot x-1}\\ f'(x) &=& \frac{\left(\frac{1}{2\cdot\sqrt{x}}+2\cdot x\right)\cdot (2\cdot x-1)-(\sqrt{x}+x^2)\cdot 2}{(2\cdot x-1)^2} \end{eqnarray}$

3: $LaTex: \begin{eqnarray} f(x) &=& (6x^4+x^2)\cdot (4\cdot\sqrt{x}-1)\\ f'(x) &=& (24x^3+2x)\cdot (4\cdot\sqrt{x}-1)+ \frac{2}{\sqrt{x}}\cdot (6x^4+x^2)\\ \end{eqnarray}$

4: $LaTex: \begin{eqnarray} f(x) &=& \frac{7}{2\cdot x}+\frac{3}{\sqrt{x}}\\ f'(x) &=& \frac{-3,5}{x^2}+\frac{-1,5}{x^{1,5}}\\ \end{eqnarray}$

5: $LaTex: \begin{eqnarray} f(x) &=& -x^4+\frac{1}{3}\cdot x^3-x^{-2}+6\\ f'(x) &=& -4\cdot x^3+x^2+\frac{2}{x^3}\\ \end{eqnarray}$

6: $LaTex: \begin{eqnarray} f(x) &=& \frac{2\cdot x -1}{x+3}\\ f'(x) &=& \frac{2\cdot (x+3)-(2\cdot x -1)\cdot 1}{(x+3)^2}\\ &=& \frac{7}{x^2+6x+9}\\ \end{eqnarray}$

7: $LaTex: \begin{eqnarray} f(x) &=& 5\cdot \frac{\sqrt{x}}{2\cdot x^3-x}\\ f'(x) &=& \frac{\frac{5}{2 \cdot \sqrt{x}} \cdot ( 2\cdot x^3 - x) - 5 \cdot \sqrt{x} \cdot (6\cdot x -1)}{(2x^3 - x)^2}\\ \end{eqnarray}$

8: $LaTex: \begin{eqnarray} f(x) &=& (-x^{-4}+\sqrt{x})\cdot (x-1)\\ f'(x) &=& \left(\frac{4}{x^5}+\frac{1}{2\sqrt{x}}\right)\cdot(x-1)+(-x^{-4}+\sqrt{x})\cdot 1\\ \end{eqnarray}$

Opgave 333

$LaTex: \begin{eqnarray} 9\cdot y &=& x-27\\ &\Downarrow&\\ y_1 &=& \frac{1}{9}\cdot x - 3\\ a_1 &=& \frac{1}{9}\\ -1 &=& a_1\cdot a_2\\ &\Downarrow&\\ a_2 &=& \frac{-1}{a_1} = -a_1^{-1} = -9\\ x_2 &=& \frac{-1}{3}\\ y_2 &=& 4\\ y_2 &=& a_2\cdot x_2 + b_2\\ &\Downarrow&\\ b_2 &=& y_2-a_2\cdot x_2 = 1\\ f_{tan}(x) &=& a_2\cdot x +b_2 = -9\cdot x + 1\\ f(x) &=& 2\cdot x^3+4\cdot x^2-7\cdot x+1\\ f_{tan}(x) &=& f(x)\\ -9\cdot x + 1 &=& 2\cdot x^3+4\cdot x^2-7\cdot x+1 0 &=& 2\cdot x^3+4\cdot x^2+2\cdot x\\ 0 &=& x\cdot(2\cdot x^2+4\cdot x+2)\\ 0 &=& 2\cdot x^2+4\cdot x+2\\ x &=& -1\\ y &=& f(x) = f_{ran}(x) = 10\\ \end{eqnarray}$

f_tan(x) berører f(x) hinanden, i punkterne x. Da f'(x) = a_2. I denne benytter vi nulreglen til at sige at enten x eller resten af formlen skal være nul i 0 = x(2x²+4x+2), derfor er resten af formlen nul.