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there's a shortcut to knowing...just look at the exponents of *every* x. What are the exponents of every x in this case?
4 & 1. would i add them? and its odd?
nope. remember: any constant is rewritten as cx^0 for example, 1 can be rewritten as 1x^0 2 can be rewritten as 2x^0 so 5 is 5x^0
so the exponents of x are actually 4, 1 and 0 got it?
yeah i get it. so its even?
now, look at 4, 1, 0 <--are these numbers even, odd or neither?
neither. because 4 even 1s odd
so this polynomial is neither
thanks, can you help me with a few more?
before that, can you give me an example of an odd function? would that be like f(x) = -5x^6 - 4x^4 - 7x^2 ?
the exponents in this case are 6, 4 and 2..are those numbers odd?
true i meant even lol so that an even function
f(x) = -5x^5 - 4x^3 - 7 woul that be odd?
it would be except for the 7, which ruins it the degree of a constant is 0, and 0 is an even number
remember, i said constants have x^0 so 7 is actually 7x^0 0 is not odd
so f(x) = -5x^5 - 4x^3 - 7x ?
or f(x) = -5x^7 - 4x^5 - 7x^3
both are odd?
yes one has exponents 7, 5, 3 <--all are odd the other is 5,3 ,1 <--all are odd
okay. so the next question. Use the graph of f to estimate the local maximum and local minimum
don't you just look at the "turning points" of that graph?
i thought so, but last tiem i got it wrong so i want to double check.
Local maximum: approx. (-1,1.17); local minimum: approx. (2,-3.33) ?
hmm...i'm not a master of graphs so i'm afraid i'll have to ask help from @helder_edwin
alright. theres one more. its graphs too tough. but i have choices, maybe that will help?
i think answers from @helder_edwin will be more precise
u r right. u got a local maximum at x=-1 and a local minimun at x=2.
alright. i have one more. can you let me know if im correct?
Determine the intervals on which the function is increasing, decreasing, and constant
i got Increasing x > 0; Decreasing x < 0
no. it is increasing over the entire real line
always read a graph from left to right.
Increasing on all real numbers?
u r welcome
@lgbasallote thanks for the referral.