Find the derivative of each function. Then, determine whether the function has any local extrema.

Question

asked Jul 20, 2023 18.9k views

1 Answer

David McLaughlin · Answer 1 · 2023-07-26T04:20:59+0000

Using the quotient rule, the derivative of
is

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

as you're shown in the picture.

If
has any local extrema, they occur at critical points where
y'=0 or possibly when
is undefined.

In this case,
is undefined when
x=1 , but this is outside the domain of
anyway, so we can ignore that.

That leaves us with the zero-case. For
$x\\eq1$ , the denominator is always positive, so the numerator must be zero. But this never happens, since

$-(x^2+1) = 0 \implies x^2 = -1$

but
$x^2\ge0$ for all real
. Therefore
has no critical points and thus no extrema.