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Question

For the function f(x)f(x) given by 2\end{cases}"\u003ef(x)={2x+4c, if x≤−1x2+5c−d, if −1x≤24x−7, if x\u003e2f(x)=\begin{cases}2x+4c,	ext{ if }x\leq-1\ $$x^2+5c-d$$,	ext{ if }-12\end{cases}, find the values of cc and dd that make it continuous everywhere.

Accepted Answer

c=0c=0d=3d=3

Answer

c=3c=3d=0d=0

Answer

c=−3c=-3﻿d=3d=3d=3﻿

Answer

﻿c=0c=0c=0﻿d=−3d=-3

For the function f(x)f(x) given by f(x)={2x+4c, if x≤−1x2+5c−d, if −1<x≤24x−7, if x>2f(x)=\(\begin{cases}\)2x+4c,\(\text{ if }\)x\(\leq\)-1\\ x^2+5c-d,\(\text{ if }\)-1<x\(\leq\)2\\ 4x-7,\(\text{ if }\)x>2\(\end{cases}\), find the values of cc and dd that make it continuous everywhere.

For the function $f (x)$ given by $f (x) = {\begin{matrix} 2 x + 4 c, if x \leq - 1 \\ x^{2} + 5 c - d, if - 1 < x \leq 2 \\ 4 x - 7, if x > 2 \end{matrix}$ , find the values of $c$ and $d$ that make it continuous everywhere.