Question

Let `f(x)=|x^2-1|/(x-1)`. Then, `f` has a local maximum at `x`=?

Let `f(x)=|x^2-1|/(x-1)`. Then, `f` has a local maximum at `x`=?

Answer 1

graph{x^2-1 [-3.465, 3.464, -1.732, 1.733]}

`abs(x^2-1) = {(x^2-1, x lt -1),(-x^2 +1, -1 le x le 1),(x^2-1, x gt 1):}`

`abs(x^2-1)/(x-1) = {(x+1, x lt -1),(-(x +1), -1 le x le 1),(x+1, x gt 1):}`

graph{abs(x^2-1)/(x-1) [-7.024, 7.02, -3.513, 3.51]}

So a local max at `x = -1`