Hi Melody!
I think: In the term \(f(f^{-1}(2010)\) is 2010 the argument of the \(f^{-1}\).
Then applies:
\(f^{-1}(x)=f^{-1}(2100)=\sqrt[3]{x}+1=\sqrt[3]{2010}+1\\ \color{blue}f^{-1}(2100)\in\{-11.6202,13.6202\}\)
These values of the function \(f^{-1}\)are two arguments for the function f.
Please confirm whether this is the right or wrong idea.
Grüße
If \(f(x)=x^3+3x^2+3x+1 \), find \(f(f^{-1}(2010)) \)
Hello Guest!
\(f(x)=x^3+3x^2+3x+1 \)
\(f(x)=y=x^3+3x^2+3x+1\\ f(x)=y=(x-1)^3\\ f^{-1}(x)=x=(y-1)^3\\ f^{-1}(x)=y=\sqrt[3]{x}+1\)
\(f^{-1}(2010)=\sqrt[3]{2010}+1=\pm 12.6202+1\\ f^{-1}(2010)\in \{13.6302,-11.6202\}\)
\(f(f^{-1}(2010))=f(13.6302, -11.6202) \)
\((x-1)^3=(13.6302-1)^3=2010\)
\((x-1)^3=(-11.6302-1)^3=-2010\)
\(f(f^{-1}(2010))\in \{-2010,2010\}\)
!
Find the sum of all values of n that satisfy 1/n-1 + 1/n+1 = 3/n
help is appreciated.
\(\color{BrickRed}1/n-1 + 1/n+1 = 3/n\\ \frac{2}{n}=\frac{3}{n}\\ n⇒\pm\infty \)
The function has no zero.
Because point calculation comes before line calculation,
you have to put the divisors in brackets.
\(1/(n-1) + 1/(n+1) = 3/n\)
\(\frac{1}{n-1}+\frac{1}{n+1}=\frac{3}{n}\\ \frac{n^2+n+n^2-n}{n(n^2-1)}=\frac{3(n^2-1)}{n(n^2-1)}\\ 2n^2=3n^2-3\\ n^2=3\\ n=\pm \sqrt{3}\)
\(n\in\{-\sqrt{3},\sqrt{3}\}\)
At the MP Donut Hole Factory, Niraek, Theo, and Akshaj are coating spherical donut holes in powdered sugar. Niraek's donut holes have radius 6 mm, Theo's donut holes have radius 10 mm, and Akshaj's donut holes have radius 18 mm. All three workers coat the surface of the donut holes at the same rate and start at the same time. Assuming that the powdered sugar coating has negligible thickness and is distributed equally on all donut holes, how many donut holes will Niraek have covered by the first time all three workers finish their current donut hole at the same time?
\(\frac{N\cdot Parts}{6^2mm^2}+\frac{T\cdot Parts}{10^2mm^2}+\frac{A\cdot Parts}{18^2mm^2}\\ =\frac{N}{36}+\frac{T}{100}+\frac{A}{324}\\ =\frac{N}{9}+\frac{T}{25}+\frac{A}{81}\\ = \frac{2025N+729T+225A}{18225}\cdot \frac{Parts}{mm^2}\)
In the period in which Akshaj 225 and Theo 729 coated donut holes,
Niraek coated 2025 donut holes.
For what values of x is \(2x^2+8x\leq -6\) ? Express your answer in interval notation.
\(2x^2+8x+6\le0\\ x^2+4x+3=0\)
p q
\(x=-\frac{p}{2}\pm \sqrt{(\frac{p}{2})^2-q}\\ x=-2\pm\sqrt{4-3}\)
\(x_1=-1\\ x_2=-3\)
\(For\ {\color{blue}-3\le x\le -1} \ is \ \color{blue}-2x^2+8x\leq -6. \)
Find the distance between \(P_1\)(3,4) and the line 4x+8y+7=0.
\(f_1(x)=y=-\frac{1}{2}x-\frac{7}{8}\)
\(m_1=-\frac{1}{2}\\ {\color{blue}m_2=-\frac{1}{m_1}}=-\frac{1}{-\frac{1}{2}}\\ \color{blue}m_2=2\)
\(\color{blue}f_2=y=m_2(x-x_p)+y_p\\ y=2(x-3)+4\\ y=2x-2\)
\(-\frac{1}{2}x-\frac{7}{8}=2x-2\\ \frac{5}{2}x=\frac{9}{8}\\ \color{blue}x_2=0.45\\ \color{blue}P_2\\ \color{blue}y_2=-1.1 \)
\(\color{blue}D=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}\\ D=\sqrt{(-1.1-4)^2+(0.45-3)^2}\\\)
\(D=5.702\)
The distance between \(P_1\)(3,4) and the line 4x+8y+7=0 is D = 5.702.
Für welche x-Werte im Intervall [−2π; 2π] gilt
sin(2x)−1,1sin(x)+0,28 = 0?
Hallo Gast, ist das deine Gleichung?
Im Intervall [−2π; 2π] gilt:
\(x\in\{-5,121; -3,323;-0,694;-0,378;\ \)
\(1,162;\ 3,051;\ 5,590;\ 5,905\}\ graphisch\ ermittelt\)
Привет друг! Значение выражения равно 4.5 ⋅ 10 − 210 − 3 = - 168
Find the length of BC, to two decimal places.
Sine law
\(\frac{8}{sin(180°-43°-61°)}=\frac{\overline{BC}}{sin(61°)}\)
\(\overline{BC}=\frac{8\cdot sin(61°)}{sin(180°-43°-61°)}\)
\(\overline{BC}=7.21\)
!
