What is the maximum number of relative extrema contained in the graph of this function: f(x)=3x^4-x^2+4x-2
hint: take the derivative of the function. then examine how many zeros that function can have, namely the max number of real zeros. that will be your answer.
Can you help?
what's the dreivative?
derivative
f(x)=12x^3-2x+4
all right. what degree polynomial is the derivative?
cubic
how many zeros will it have?
10?
really? have you heard of the fundamental theorem of algebra?
Yeah? it has at least one complex root
http://www.mathsisfun.com/algebra/fundamental-theorem-algebra.html have a look at this...
so 3
@pgpilot326 where'd you go!
right here... great. does that make sense?
Yeah,so how do I find max number of relative extrema
each relative extrema will occur at a zero of the derivative function. so if your function is a polynomial then the max number will be 1 less than it's degree.
So the max number is 3?
yep and the minimum number is 1. that's because when a polynomial has real coefficients, complex roots occur in conjugate pairs. since 3 is the max, 3-2 = 1 would be the min. (number of relative extrema won't be below 0).
Ok that makes a lot of sense! Would you mind helping on one more? It is pretty easy identifying a polynomial from a multiple choice list. Just be check my answer
sure
WHich is a polynomial: f(x)=4/x - 11x^2 f(x)= -2.1^7+3x^2+17/2 x ---- my answer f(x)= -1/5x^2 +4.5x + 2^x f(x)= -3x^4+8x^2-2sqrtx + 1
a polynomial must have non-negative, integer exponents.
Ok my answer is correct then
so in your answer, is the last term \[\frac{ 17 }{ 2x }\text{ or }\frac{ 17 }{ 2 }x\]
the second one
good
Thanks!
you're welcome!
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