79997
Let \(f(x)=\left\{\begin{array}{c}|x|,-\infty\lt x\lt 2 \\ |2 x-4|, 2 \leq x \leq 20\end{array}\right.\)
\(x=a\) is a point where \(f(x)\) is continuous but not differentiable and \(x=b\) is a point where \(f(x)\) is not differentiable \((a \neq b)\). Then \(a+b=\)
79999
\(f(x)=\left\{\begin{array}{cc}\frac{x-|x|}{x}, & \text { when } x\lt 0 \\ 5 x^2+a, & \text { when } 0 \leq x \leq 1 \\ b\left(\frac{x^2-1}{x^2-3 x+2}\right), & \text { when } 1\lt x\lt 3 \\ -14, & \text { when } x>3\end{array}\right.\) is a
continuous function on \(R\), the \((a, b)\) =
79997
Let \(f(x)=\left\{\begin{array}{c}|x|,-\infty\lt x\lt 2 \\ |2 x-4|, 2 \leq x \leq 20\end{array}\right.\)
\(x=a\) is a point where \(f(x)\) is continuous but not differentiable and \(x=b\) is a point where \(f(x)\) is not differentiable \((a \neq b)\). Then \(a+b=\)
79999
\(f(x)=\left\{\begin{array}{cc}\frac{x-|x|}{x}, & \text { when } x\lt 0 \\ 5 x^2+a, & \text { when } 0 \leq x \leq 1 \\ b\left(\frac{x^2-1}{x^2-3 x+2}\right), & \text { when } 1\lt x\lt 3 \\ -14, & \text { when } x>3\end{array}\right.\) is a
continuous function on \(R\), the \((a, b)\) =
79997
Let \(f(x)=\left\{\begin{array}{c}|x|,-\infty\lt x\lt 2 \\ |2 x-4|, 2 \leq x \leq 20\end{array}\right.\)
\(x=a\) is a point where \(f(x)\) is continuous but not differentiable and \(x=b\) is a point where \(f(x)\) is not differentiable \((a \neq b)\). Then \(a+b=\)
79999
\(f(x)=\left\{\begin{array}{cc}\frac{x-|x|}{x}, & \text { when } x\lt 0 \\ 5 x^2+a, & \text { when } 0 \leq x \leq 1 \\ b\left(\frac{x^2-1}{x^2-3 x+2}\right), & \text { when } 1\lt x\lt 3 \\ -14, & \text { when } x>3\end{array}\right.\) is a
continuous function on \(R\), the \((a, b)\) =
79997
Let \(f(x)=\left\{\begin{array}{c}|x|,-\infty\lt x\lt 2 \\ |2 x-4|, 2 \leq x \leq 20\end{array}\right.\)
\(x=a\) is a point where \(f(x)\) is continuous but not differentiable and \(x=b\) is a point where \(f(x)\) is not differentiable \((a \neq b)\). Then \(a+b=\)
79999
\(f(x)=\left\{\begin{array}{cc}\frac{x-|x|}{x}, & \text { when } x\lt 0 \\ 5 x^2+a, & \text { when } 0 \leq x \leq 1 \\ b\left(\frac{x^2-1}{x^2-3 x+2}\right), & \text { when } 1\lt x\lt 3 \\ -14, & \text { when } x>3\end{array}\right.\) is a
continuous function on \(R\), the \((a, b)\) =