Find the t-coordinates of all critical points of the given function. Determine whether each critical point is a relative maximum, minimum, or neither by first applying the second derivative test, and, if the test fails, by some other method.

f(t)= -3t^3+2t

a.) f has (a relative maximum, relative minimum, or no relative extrema) at the critical point t=________ * (smaller t-value)

b.) f has (a relative maximum, relative minimum, or no relative extrema) at the crtical point t=__________* (larger t-value)

Respuesta :

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]

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