Find the point(s) on the graph of f where the tangent line is horizontal.

Question

f(x) = x / (x^2 + 36)

Answer 1

The function f(x) = x/(x² + 36)

y = x/(x² + 36)

Apply derivative on each side with respect of x.

Apply quotient rule in derivatives d/dx (u/v) = [vu' - uv']/v²

u = x, v = x² + 36

u' = 1, v' = 2x

y' = [(x² + 36)(1) - (x)(2x)]/(x² + 36)²

= [(x² + 36) - 2x² ]/(x² + 36)²

= [ - x² + 36 ]/ (x² + 36)²

Set first derivative equals to zero.

[ - x² + 36 ]/ (x² + 36)² = 0

- x² + 36 = 0

x² = 36

Apply square root on each side.

x = ± 6

Substitute x = ± 6 in  y = x/(x² + 36).

For x = 6

y = x/(x² + 36)

y = 6/(6² + 36)

y = 6/72

y = 0.083

For x = - 6

y = (-6)/[(-6)² + 36)]

y = - 6/72

y = - 0.083

The tangent is horizontal at two points are (6, 0.083) and ( -6, -0.083).