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can you give me the derivative of f ??
that's all it says :// :(
yes... but you need to calculate the derivative of f to find the relative max/min of the function.
isn't this for a calculus class?
its okay :) so you don't know how to do it? :(
i know how to do it on a calculater ..but i left it at a friends house ://
oh... so you're allowed to use the max/min functions of a graphing calculator?
ok... hang on...
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:
it looks like you have a relative max at x=-4 and a relative min at x=0
yeah, but that's not one of my options :(
the function you gave at the start does not coincide with those choices...
thats what it says in my homework
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.
f(-4) = -4 so (-4, -4) is a relmax f(0) = -36 so (0, -36) is a relmin
but that's none of the options ://
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?
yes it is ..hmm :// :(
im just going to pick a or b:///.. lol ..thank you for the help though :)
idk what to say... :(
do you think you could help me with another one?
there has to be an error in those choices... yw... :)
What is a cubic polynomial function in standard form with zeros 1, –2, and 2?
do you know how to multiply out binomials?
i know i have to write like (x+1)(x-2)(x+2)
standard form for a cubic polynomial is: \(\large y=ax^3+bx^2+cx+d \) where a, b, c, d are real numbers.
not quite... since the zeros are 1, -2, and 2, then the factors are (x-1)(x+2)(x-2)
so all you do is multiply out the 3 binomials....
how is it -1 when it says 1 ?
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
multiply out the binomials: (x-1)(x+2)(x-2) = \(\large x^3-x^2-4x+4 \)