# Show that function f(x)=2x^3+6x-5 has not any extremes!

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At the extreme values of a function the first derivative is 0. Or to find the extreme value for f(x), find f'(x), equate it to 0 and solve for x.

f(x) = 2x^3 + 6x - 5

f'(x) = 6x^2 + 6

6x^2 + 6 = 0

=> x^2 = -1

**This only gives a complex root for x. As extreme values can be found only if the value of x is real, the given function does not have any extremes.**

For a function to allow a local extreme , it's derivative has to have real roots.

We'll determine the derivative of the function:

f'(x) = 6x^2 + 6

6x^2 + 6 > 0, for any real x

It is obvious that the equation is not cancelling for any real value of x, so the 1st derivative has no real roots.

**Since the derivative of the function is strictly positive, the function does not allow local extremes.**