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

Asked on by orlovolga

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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.

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

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