Answer:
f has relative maximum at t = [tex]+\frac{\sqrt2}{3}[/tex]
and
f has relative minimum at t = [tex]-\frac{\sqrt2}{3}[/tex]
Step-by-step explanation:
Data provided in the question:
f(t) = -3t³ + 2t
Now,
To find the points of maxima or minima, differentiating with respect to t and putting it equals to zero
thus,
f'(t) = (3)(-3t²) + 2 = 0
or
-9t² + 2 = 0
or
t² = [tex]\frac{2}{9}[/tex]
or
t = [tex]\pm\frac{\sqrt2}{3}[/tex]
to check for maxima or minima, again differentiating with respect to t
f''(t) = 2(-9t) + 0 = -18t
substituting the value of t
at t = [tex]+\frac{\sqrt2}{3}[/tex]
f''(t) = [tex](-18)\times\frac{\sqrt2}{3}[/tex]
= - 6√2 < 0 i.e maxima
and at t = [tex]-\frac{\sqrt2}{3}[/tex]
f''(t) = [tex](-18)\times(-\frac{\sqrt2}{3})[/tex]
= 6√2 > 0 i.e minima
Hence,
f has relative maximum at t = [tex]+\frac{\sqrt2}{3}[/tex]
and
f has relative minimum at t = [tex]-\frac{\sqrt2}{3}[/tex]