how many local extremes for x^4-2x^3-8x+16. the answer says its just one local extremes, but why not 3. since it's 4th degree?

Question



Answer 1

Well, since this is precalculus, we can't really talk about derivatives yet. Your only available good tool is graphing, like the other guy said.

In this particular case you can also check by factoring the expression. This is not always very helpful.

x^4-2x^3-8x+16 = (x³-8)(x - 2) = (x-2)²(x² + 2x +4) where (x-2) gives you the double root and extremum at 2; and (x² + 2x + 4) is an irreducible quadratic with no roots

Simply put, polynomials of degree n can have n-1 local extrema or less. Think of x⁴ having only one minimum, or x³ having none at all.

Answer 2

The easiest way to see this is to graph it.

You will see that even though this is a quartic (4th degree) is has only one minimum point at x = 2.

At the minimum the the value of the expression is 0, that means the derivative there is 0 and, therefore, 2 is a double root. If you divide the expression by (x² - 4x + 4), you will find that the other two roots are complex.