Continuity by Definition – Continuity check by definition to a function with a parameter – Exercise 891 Post category:Continuity by Definition Post comments:0 Comments Exercise Given the function (c parameter) f(x)={2x−1x,x≠0c,x=0f(x) = \begin{cases} \frac{2^x-1}{x}, &\quad x\neq 0 \\ c, &\quad x =0\\ \end{cases}f(x)={x2x−1,c,x=0x=0 For which values of the parameter is the function continuous? Final Answer Show final answer c=ln2c=\ln 2c=ln2 Solution Coming soon… Share with Friends Read more articles Previous PostContinuity by Definition – Continuity check by definition to a function with parameters – Exercise 898 Next PostContinuity by Definition – Continuity check by definition to a function with a parameter – Exercise 884 You Might Also Like Proof of Continuity – A split function with a function to the power of a function – Exercise 6236 July 5, 2019 Proof of Continuity – A split function with a rational function and a parameter – Exercise 6252 July 5, 2019 Proof of Continuity – A split function with ln and a third root – Exercise 6240 July 5, 2019 Proof of Continuity – A split function with exponential functions and a parameter – Exercise 6591 July 16, 2019 Proof of Continuity – A split function with exponential functions with a parameter – Exercise 6248 July 5, 2019 Proof of Continuity – A split function with polynomial functions and a parameter – Exercise 5876 June 30, 2019 Leave a Reply Cancel replyCommentEnter your name or username to comment Enter your email address to comment Enter your website URL (optional) Δ
Proof of Continuity – A split function with a function to the power of a function – Exercise 6236 July 5, 2019
Proof of Continuity – A split function with a rational function and a parameter – Exercise 6252 July 5, 2019
Proof of Continuity – A split function with exponential functions and a parameter – Exercise 6591 July 16, 2019
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