Practicing Success
Practicing Success
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If x + 2|y| = 3y, where y = f(x), then f(x) is |
continuous everywhere differentiable everywhere discontinuous at x = 0 none of the above |
continuous everywhere |
$\left.\begin{matrix}x+2y=3y&(y≥0)\\x-2y=3y&(y<0)\end{matrix}\right\}⇒y=\left\{\begin{matrix}x&(x≥0)\\\frac{x}{5}&(x<0)\end{matrix}\right.$ function is continuous everywhere, but not differentiable at x = 0. |
