asinus

Excuse me, I have mistakenly entered this function:

\(f(x)=e^x-3e^{-x}>0\)

Here the graph of the correct function:

\(e^x-3e^{-x}+2>0\)

Guest two has right

\(e^x-1>0\)

That is whenever \(x>0\)

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\(\)

Hi, Melody

my graphing calculator is

what is 40/7

\(\frac{40}{7}=5\frac{5}{7}=5.71428\overline{571428} \ periodic\)

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sorry,I typed wrong. 

(x/8)-(x/2)=6+(6x/16)

\(\frac{x}{8}-\frac{x}{2}=6+\frac{6x}{16} \)

\(\frac{2x-8x}{16} =\frac{16* 6+6\color{red}x}{16} \)  

\(-6x-6x=96\)

\(-12x=96\)

\(x=-8\)     !

(x/8)-(x/2)=6+(6x/16)

\(\frac{x}{8}-\frac{x}{2}=6+\frac{6x}{16} \)    [least common multiple = 16

\(\frac{2x-8x}{16} =\frac{16* 6+6}{16} \)       [ add, subtr, multiply 16 

\(-6x=102\)               [ / - 6

\(x=-17\)     !

e^x-3e^(-x)+2>0 {nl} how can i solve this disequation?

\(f(x)=e^x-3e^{-x}+2>0\)

\(x\geq 0,54933\)   (laut Wertetabelle MatheGrafix)

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You're soo right, Melody!
There are two minus signs. 

3y--7y=

\(3y-7y=-4y \)     !

2=lnx+2x the answer is x=1 but i dont see who they got that

\(2=lnx+2x\)

\(lnx=2(1-x)\)

\(\frac{lnx}{2}=1-x\)

\(ln\sqrt x=1-x\)

\(x+ln\sqrt x=1\)

\(f(x)=x+ln\sqrt x-1=0\)

\(\LARGE x=1\)

  !

26+x(3+4x)

\(26+x(3+4x)=26+3x+4x^2\)    !