Where is the function #h(x)=abs(x-1)+abs(x+2)# differentiable?
It is differentiable everywhere except the zeros of the expressions in absolute value. Everywhere except
The short answer statement is correct. Functions incolving terms that have absolute value tend to be non-differentiable at the zeros of the expression in absolute value. This is not true if the expression as always non-negative or always non-positive.
But let's analyze this function.
Combining, we have
Differentiating, we get
Because the derivatives do not agree at the joints (the left and right derivatives are distinct), the function is not differentiable at the joints.
graph{abs(x-1)+abs(x+2) [-11.235, 11.265, -1.74, 9.51]}
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The function ( h(x) = |x - 1| + |x + 2| ) is differentiable everywhere except at the points where the absolute value function changes direction abruptly. These points are ( x = -2 ) and ( x = 1 ).
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The function h(x) = |x - 1| + |x + 2| is differentiable at all points where it is continuous. Since |x - 1| and |x + 2| are both continuous everywhere, h(x) is continuous everywhere. Therefore, h(x) is differentiable for all real numbers x.
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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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