MATH SOLVE

4 months ago

Q:
# 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+2ta.) 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)

Accepted Solution

A:

Answer:f has relative maximum at t = [tex]+\frac{\sqrt2}{3}[/tex]andf has relative minimum at t = [tex]-\frac{\sqrt2}{3}[/tex]Step-by-step explanation:Data provided in the question:f(t) = -3t³ + 2tNow,To find the points of maxima or minima, differentiating with respect to t and putting it equals to zerothus,f'(t) = (3)(-3t²) + 2 = 0or-9t² + 2 = 0ort² = [tex]\frac{2}{9}[/tex]ort = [tex]\pm\frac{\sqrt2}{3}[/tex]to check for maxima or minima, again differentiating with respect to tf''(t) = 2(-9t) + 0 = -18tsubstituting the value of tat t = [tex]+\frac{\sqrt2}{3}[/tex]f''(t) = [tex](-18)\times\frac{\sqrt2}{3}[/tex]= - 6√2 < 0 i.e maximaand at t = [tex]-\frac{\sqrt2}{3}[/tex]f''(t) = [tex](-18)\times(-\frac{\sqrt2}{3})[/tex] = 6√2 > 0 i.e minimaHence,f has relative maximum at t = [tex]+\frac{\sqrt2}{3}[/tex]andf has relative minimum at t = [tex]-\frac{\sqrt2}{3}[/tex]