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GO Classes DPP | ML |GRADIENT

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2 2 votes

Consider the following piece-wise function as shown in the image:

$$
y(x)= \begin{cases}x^2 & x \leq 0 \\ 2 x & 0 \leq x \leq 2 \\ 4 x-4 & 2 \leq x\end{cases}
$$


How many sub-gradients are possible at points $a_1, a_2, a_3$ and $a_4$ ?

  1. $a_1$ : Many, $a_2$ : One, $a_3$ : Many, $a_4$ : One
     
  2. $a_1$ : One, $a_2$ : Many, $a_3$ : Many, $a_4$ : One
     
  3. $a_1$ : One, $a_2$ : Many, $a_3$ : One, $a_4$ : Many
     
  4. $a_1$ : Many, $a_2$ : One, $a_3$ : One, $a_4$ : Many

1 Answer

1 1 vote
$a_1$ lies on the part of the function which is differentiable. For a differentiable function (subpart), only one sub-gradient is possible which is the gradient itself.

$a_2$ lies at the intersection of two $x^2$ and $2 x$. The function is not differentiable at this point (as left slope is different from right slope). Hence there are multiple sub-gradients possible at $a_2$.

$a_3$ lies on the part of the function which is differentiable. For a differentiable function (subpart), only one sub-gradient is possible which is the gradient itself.

$a_4$ lies at the intersection of two $2 x$ and $4 x-4$. The function is not differentiable at this point (as left slope is different from right slope). Hence there are multiple sub-gradients possible at $a_2$
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