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dpaInc
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can you give me the derivative of f ??
heathernelly
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that's all it says :// :(
dpaInc
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yes... but you need to calculate the derivative of f to find the relative max/min of the function.
dpaInc
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isn't this for a calculus class?
heathernelly
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algebra 2
dpaInc
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oh... sorry....
heathernelly
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its okay :) so you don't know how to do it? :(
heathernelly
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i know how to do it on a calculater ..but i left it at a friends house ://
dpaInc
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oh... so you're allowed to use the max/min functions of a graphing calculator?
heathernelly
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yes
dpaInc
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ok... hang on...
heathernelly
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okaay :)
dpaInc
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since you don't have your graphing calculator with you, use this online one....
https://www.desmos.com/calculator
i've also included the graph of your function here:
dpaInc
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it looks like you have a relative max at x=-4 and a relative min at x=0
heathernelly
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yeah, but that's not one of my options :(
dpaInc
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the function you gave at the start does not coincide with those choices...
heathernelly
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thats what it says in my homework
dpaInc
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i double checked with this function you gave at the start: \(\large f(x)=x^3+6x^2-36 \)
f has a relative max at x=-4 and a relative min at x=0.
dpaInc
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f(-4) = -4 so (-4, -4) is a relmax
f(0) = -36 so (0, -36) is a relmin
heathernelly
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but that's none of the options ://
dpaInc
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i'm sorry.... i'm absolutely sure of my answer here..
even the calculator verifies my answer....
is this the function we started with correct?
heathernelly
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yes it is ..hmm :// :(
heathernelly
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im just going to pick a or b:///.. lol ..thank you for the help though :)
dpaInc
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idk what to say... :(
heathernelly
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do you think you could help me with another one?
dpaInc
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there has to be an error in those choices...
yw... :)
dpaInc
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yes...
heathernelly
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What is a cubic polynomial function in standard form with zeros 1, –2, and 2?
dpaInc
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do you know how to multiply out binomials?
heathernelly
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i know i have to write like (x+1)(x-2)(x+2)
dpaInc
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standard form for a cubic polynomial is: \(\large y=ax^3+bx^2+cx+d \)
where a, b, c, d are real numbers.
dpaInc
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not quite... since the zeros are 1, -2, and 2,
then the factors are (x-1)(x+2)(x-2)
dpaInc
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so all you do is multiply out the 3 binomials....
heathernelly
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how is it -1 when it says 1 ?
dpaInc
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a zero refers to where the graph crosses the x-axis (when y=0)
so the equation is 0 = (x-1)(x+2)(x-2)
if x=1, you'll have 0 = (1-1)(1+2)(1-2) which is a true statement
but if you have what you wrote down, 0 = (1+1)(1+2)(1-2) is not a true statement
dpaInc
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multiply out the binomials: (x-1)(x+2)(x-2) = \(\large x^3-x^2-4x+4 \)
heathernelly
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thanks :)