Function Posted November 24, 2014 Share Posted November 24, 2014 (edited) Hi everyone Out of boredom, I asked myself the following question: For which [math]a[/math], [math]b[/math], [math]c[/math], [math]d[/math], [math]e[/math] and [math]f[/math] does the curve, described by the function [math]g(x)=a\cdot\log_b{\left(c\cdot x^d+e\right)}+f[/math] Only have one solution, so touches the curve described by the function [math]h(x)=x[/math] In [math](1,1)[/math]? Found something, and wanted to know if it was right. It's been a while since I 'performed' pure mathematics (over 4 months), so don't be too hard on me The derivative of that function in [math]x=1[/math] should be 1. [math]\frac{d}{dx}a\cdot\log_b{\left(c\cdot x^d+e\right)}+f = 1[/math] [math]\Leftrightarrow a\cdot\frac{d}{dx}\left(\log_b{\left(c\cdot x^d+e\right)}\right) = 1[/math] [math]\Leftrightarrow a\cdot \frac{c\cdot d\cdot x^{d-1}}{\left(c\cdot x^d+e\right)\cdot\ln{b}} = 1[/math] (If there's a mistake, it's in the equation above this line... forgot how the derivative of a logarithmic function, so had to do it manually and found that the derivative of y to x of the function y = log_b{x} is 1/(xlnb)) [math]\Leftrightarrow a\cdot c\cdot d\cdot x^{d-1} = \left(c\cdot x^d+e\right)\cdot\ln{b}[/math] [math]x = 1[/math] [math]\Leftrightarrow a\cdot c\cdot d = (c+e)\cdot\ln{b}[/math] Second possible expression: [math]a\cdot\log_b{\left(c+e\right)}+f=1[/math] Are these valid expressions to get the wanted? Thanks. F. Edited November 24, 2014 by Function Link to comment Share on other sites More sharing options...
elfmotat Posted November 24, 2014 Share Posted November 24, 2014 I'm not sure what you mean by "only have one function, so touches the curve described by the function." Could you explain? Link to comment Share on other sites More sharing options...
Function Posted November 24, 2014 Author Share Posted November 24, 2014 I'm not sure what you mean by "only have one function, so touches the curve described by the function." Could you explain? replaced 'function' with 'solution' so that the tangent in x = 1 of that function is the function x. Link to comment Share on other sites More sharing options...
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