Question

Find any relative extrema ?(?) = 3?^4 − 2?^3 + 4

Answer 1

given,

then,

for critical points,

x = 0 and x = 6/12 = 0.5

again derivative of it,

at x = 0,

g''(x) = 36*0 - 12*0 = 0

So, x = 0 is a saddle point.

at x = 0.5

g''(x) = 36*(0.5^2) - 12*(0.5) = 3

Since, at x = 0.5, g''(x) < 0 So, x = 0.5 is extrema minimum point.

So, at x = 0.5

g(0.5) = 3*(0.5^4) - 2*(0.5^3) + 4 = 3.9375

then, Relative extrema of g(x) = (0.5 , 3.9375)

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