Given #f(t) = (t^2-1)/t#: What are the stationary points of #f(t)# on the real co-ordinate plane, if any?

1 Answer
Nov 11, 2017

No stationary points on the real co-ordinate plane.

Explanation:

#f(t) = (t^2-1)/t#

Apply quotient rule

#f'(t) = ((t*2t)-(t^2-1)*1)/t^2#

#= (2t^2-t^2+1)/t^2 = (t^2+1)/t^2#

For stationary points #f'(t) =0#

Thus, where: # (t^2+1)/t^2 =0#

#-> t^2+1 = 0#

No real values of #t# will satisfy the above equation (#t=+-i#)

Hence, #f(t)# has no stationary points on the real co-ordinate plane

We can deduce this result from the graph of #f(t)# below.

graph{(x^2-1)/x [-7.023, 7.024, -3.51, 3.513]}