Question

Derivative of the absolute value of x-1?

Answer 1

what is the derivative of \(-x+1\)?

Answer 2

-1

Answer 3

derive implicitly

y = |x-1|
y^2 = (x-1)^2
2y y' = 2(x-1)
y' = 2(x-1)/(2y) = (x-1)/|x-1|

Answer 4

and what is the derivative of \(x-1\) ?

Answer 5

1

Answer 6

\[f(x) = |x-1| = \left\{\begin{array}{rcc}
-x+1& \text{if} & x <1 \\
x-1& \text{if} & x \geq 14\end{array}
\right.
\]

Answer 7

typo there but you get the idea right?

Answer 8

I think so. from there how would you get the value of x on the clos interval 0 to 4?

Answer 9

\[f(x) = |x-1| = \left\{\begin{array}{rcc}
-x+1& \text{if} & x <1 \\
x-1& \text{if} & x \geq 1\end{array}
\right.\]
\[f(x) = |x-1| = \left\{\begin{array}{rcc}
-1& \text{if} & x <1 \\
1& \text{if} & x \geq 1\end{array}
\right.\]

Answer 10

that should have been
\[f'(x) = |x-1| = \left\{\begin{array}{rcc}
-1& \text{if} & x <1 \\
1& \text{if} & x > 1\end{array}
\right.\]

Answer 11

also \(f'(1)\) does not exist as \(1\neq -1\)

Answer 12

i am not sure what you mean by "the value of x on the closed interval"