A line with slope m passes through the origin and is tangent
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BigGlenntheHeavy
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Re: help...
\(\displaystyle f(x) \ = \ y \ = \ ln(2x) \ \implies \ y' \ = \ \frac{1}{x}\)
Let the point on f(x) that is tangent to the line emanating from the point (0,0) be (x,y).
\(\displaystyle Hence, y-0 \ = \ \frac{1}{x}(x-0) \ or \ y \ = \ 1.\)
\(\displaystyle Ergo, \ y \ = \ 1 \ and \ y \ = \ ln(2x), \ \implies \ x \ = \ \frac{e}{2}.\)
\(\displaystyle This \ implies \ that \ m \ = \ \frac{2}{e}.\)
\(\displaystyle Then, \ y-0 \ = \ \frac{2}{e}(x-0) \ = \ \frac{2x}{e}, \ see \ graph \ below.\)
[attachment=0:3erjrixr]even.jpg[/attachment:3erjrixr]
\(\displaystyle f(x) \ = \ y \ = \ ln(2x) \ \implies \ y' \ = \ \frac{1}{x}\)
Let the point on f(x) that is tangent to the line emanating from the point (0,0) be (x,y).
\(\displaystyle Hence, y-0 \ = \ \frac{1}{x}(x-0) \ or \ y \ = \ 1.\)
\(\displaystyle Ergo, \ y \ = \ 1 \ and \ y \ = \ ln(2x), \ \implies \ x \ = \ \frac{e}{2}.\)
\(\displaystyle This \ implies \ that \ m \ = \ \frac{2}{e}.\)
\(\displaystyle Then, \ y-0 \ = \ \frac{2}{e}(x-0) \ = \ \frac{2x}{e}, \ see \ graph \ below.\)
[attachment=0:3erjrixr]even.jpg[/attachment:3erjrixr]
