How do you find the exact relative maximum and minimum of the polynomial function of #f(x) = x^3 + x + 4/x#?

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Answer 1 · 2018-04-12T05:20:34

# f_min=f(1)=6, and, f_max=f(-1)=-6#.

We know that, for # (1) :f_max, f'(x)=0 and f''(x) lt 0, and, #

# (2) : f_min, f'(x)=0, and, f''(x) gt 0#.

Now, #f(x)=x^3+x+4/x#,

# rArr f'(x)=3x^2+1-4/x^2, and, f''(x)=6x+8/x^3#.

#:. f'(x)=0 rArr 3x^2+1-4/x^2=0, or, 3x^4+x^2-4=0#.

#:. (x^2-1)(3x^2+4)=0#.

#because AA x in RR, 3x^2+4 !=0, x^2-1=0, or, x=+-1#.

Then, #f''(1)=6+8=14 gt 0, f_min=f(1)=6#.

Further, #f''(-1)=-14 lt 0, f_max=f(-1)=-6#.

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