We have #f:RR->RR,f(x)=root(3)(x^2+mx+1)#.Which are values of #m# for which the graph of #f# has tangent in every point in #RR#?

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Answer 1 · 2018-03-12T12:28:38

#-2 < m < 2#

The derivative of the function:

#f(x) = root(3)(x^2+mx+1) = (x^2+mx+1)^(1/3)#

is:

#f'(x) = 1/3 (2x+m)(x^2+mx+1)^(-2/3) = (2x+m)/(3root(3)((x^2+mx+1)^2)#

For the tangent to exist for every #x in RR# the denominator of #f'(x)# must never vanish, so the polynomial:

#x^2+mx+1#

must have no real roots, that is its determinant:

#Delta = m^2-4#

must be negative.

Solving the inequality:

#m^2 - 4 < 0#

we then find:

#-2 < m < 2#