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You need to evaluate the monotony of the function, hence, you need to remember that the function increases if f'(x)>0 and the function decreases if f'(x)<0.
You need to evaluate the first derivative of the function:
`f'(x) = (ln(x^4+27))' `
`f'(x) = (1/(x^4+27))*(x^4+27)'`
`f'(x) = (4x^3)/(x^4+27)`
You need to set f'(x) = 0:
`(4x^3)/(x^4+27) = 0`
`4x^3 = 0 => x = 0`
You need to notice that f'(x)>0 for `x in (0,+oo)` and f'(x)<0 for `x in (-oo,0), ` hence, the function increases for `x in (0,+oo)` and it decreases for `x in (-oo,0).`
b) The local maximum and minimum values are those x values for f'(x) = 0. From previous point a) yields that f'(x) = 0 for x = 0 and the function decreases as x approaches to 0, from the left, and then it increases.
Hence, the function has only a minimum point at x = 0, and the point is (0, ln27).
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