Proving Derivative Existence – A function with parameters – Exercise 1231 Post category:Proving Derivative Existence Post comments:0 Comments Exercise Given the function (a and b parameters) f(x)={ax+1,0≤x≤35−bx,3<x≤6f(x) = \begin{cases} a\sqrt{x+1}, &\quad 0\leq x\leq 3 \\ 5-bx, &\quad 3<x\leq 6\\ \end{cases}f(x)={ax+1,5−bx,0≤x≤33<x≤6 For which values of the function parameters is it differentiable at point x=3? Final Answer Show final answer a=4,b=−1a=4, b=-1a=4,b=−1 Solution Coming soon… Share with Friends Read more articles Next PostProving Derivative Existence – A polynomial and an exponential functions – Exercise 1150 You Might Also Like Proving Derivative Existence – A multiplication with sin function – Exercise 1094 December 10, 2018 Proving Derivative Existence – A multiplication with sin function – Exercise 1101 December 10, 2018 Proving Derivative Existence – A function with parameters – Exercise 1123 December 12, 2018 Proving Derivative Existence – A function with parameters – Exercise 1132 December 13, 2018 Proving Derivative Existence – A function with a polynomial and a square root – Exercise 1140 December 13, 2018 Proving Derivative Existence – A polynomial function inside a square root – Exercise 1147 December 13, 2018 Leave a Reply Cancel replyCommentEnter your name or username to comment Enter your email address to comment Enter your website URL (optional) Δ
Proving Derivative Existence – A function with a polynomial and a square root – Exercise 1140 December 13, 2018
Proving Derivative Existence – A polynomial function inside a square root – Exercise 1147 December 13, 2018