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Andrei [34K]
1 year ago
8

This function Is it differentiable at x = 1​ ?

Mathematics
2 answers:
nikklg [1K]1 year ago
8 0

It is not differentiable at x=1 since the slope of the tangent line as x -> 1 from the right is 1 while the slope of the tangent line as x->1 from the left is -1

Reptile [31]1 year ago
4 0

By definition of absolute value:

|x - 1| = \begin{cases} x - 1 & \text{if } x \ge 1 \\ -(x-1) & \text{if } x < 1 \end{cases}

Then the derivative is

\dfrac{d|x-1|}{dx} = \begin{cases} 1 & \text{if } x > 1 \\ -1 & \text{if } x < 1\end{cases}

but does not exist at x=1 because

\displaystyle \lim_{x\to1^-} f'(x) = -1

while

\displaystyle \lim_{x\to1^+} f'(x) = 1

and these limits are not equal, so the derivative is discontinuous and hence |x-1| is not differentiable at x=1.

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<u>Step-by-step explanation:</u>

Isolate w by performing the following steps

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y=\dfrac{1}{2}+\dfrac{w}{3}\\\\\\6\bigg[y=\dfrac{1}{2}+\dfrac{w}{3}\bigg]\quad \implies \quad 6y=3+2w\\\\\\6y-3=3-3+2w\quad \implies \quad 6y-3=2w\\\\\\\dfrac{6y-3}{2}=\dfrac{2w}{2}\quad \implies \quad \large\boxed{\dfrac{6y-3}{2}=w}

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You can solve for x by subtracting 26 from both sides of the equation.

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