Answers
f(x) = 0.0012 +4.2x-10.
F'(x) = 0.001 (2x) + 4.2 f'(x) = 0.002x+4 2. solve f'(x) =0 for coritical values. 0.0022+4.2-0. 0.002X = -4-20 -4.2 0.002 x = -2100. a- - 2100 is the critical value is v f"(*) - - .
(f/m) Foto (0.002x+4+2) = 0.002 P" () f"(-2100) -0-002 If f (x0) 70 then f(x) has a minimum at nako Here f"(-2100) 70 So f(x) has a minimum at a=-2100.
The absolute minimum is at x=-2100 and there is no absolute maximum
.
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