Questions: Let f be defined by f(x)= 4x+m, x<2 -6x^2+2m, x >= 2 (a) Find (in terms of m ): lim x->2+ f(x)=2m-24 (b) Find (in terms of m ): lim x->2- f(x)=m+8 (c) Find all values of m such that lim x->2- f(x)=lim x->2+ f(x) Values of m= DNE

Solution Steps

For \(x \geq 2\), the function is defined as \(f(x) = -6x^{2} + 2m\). To find \(\lim _{x \rightarrow 2^{+}} f(x)\), substitute \(x = 2\) into the expression:
\[ \lim _{x \rightarrow 2^{+}} f(x) = -6(2)^{2} + 2m = -24 + 2m. \]

For \(x < 2\), the function is defined as \(f(x) = 4x + m\). To find \(\lim _{x \rightarrow 2^{-}} f(x)\), substitute \(x = 2\) into the expression:
\[ \lim _{x \rightarrow 2^{-}} f(x) = 4(2) + m = 8 + m. \]

For the limits to be equal, we set \(\lim _{x \rightarrow 2^{-}} f(x) = \lim _{x \rightarrow 2^{+}} f(x)\):
\[ 8 + m = -24 + 2m. \]
Subtract \(m\) from both sides:
\[ 8 = -24 + m. \]
Add \(24\) to both sides:
\[ m = 32. \]
Thus, the value of \(m\) that satisfies the condition is \(m = 32\).

Final Answer