This is not trigonmetry. You may have to do some plotting of G(x). We cannot do the graphing for you.
The relative extreme values can be obtained by setting the derivative equal to zero, giving
3 x^2 = 8
and solving for x
Please show your work if you need further assistance.
use the function G(x)=x^3-8x +1
b) identify and verify the y intercept of G
c) estimate the zeros of G to the nearest tenth
d) estimate the coordinates of the relative extrema of G to the nearest hundredth
e)in what intervals is G positive? Negative?
f) in what intervals is G increasing? Decreasing?
4 answers
this is trigonometry.. its the class i'm taking and my book is called, functions, statistics, and trigonometry. scott, foresman.
when x = 0, G = 1
sketch a graph of this thing right off. Let y = G(x)
when x is big negative, y is big negative
when x is big positive, y is big positive
we know when x = 0, y = 1
since it is cubic, it may cross the x axis three times
when x = -1, y = 8
when x = +1, y = -6
That gives you a pretty good idea what it looks like
there is a zero left of -1, a zero between 0 and +1, and a zero right of +1
I will leave finding those zeros for you, guess and then guess again
Now You know there is an max around -1 and a min around +1
I would take the derivative to look for where, but you can guess and correct the guess
taking derivative and setting to zero
0 = 3 x^2 -8
x^2 = 8/3
x = +/- 1.63
so we have a max at x = -1.63 , you find y
we have a min at x=+1.63, you find y
Now you can easily make your graph more accurate to answer e and f
sketch a graph of this thing right off. Let y = G(x)
when x is big negative, y is big negative
when x is big positive, y is big positive
we know when x = 0, y = 1
since it is cubic, it may cross the x axis three times
when x = -1, y = 8
when x = +1, y = -6
That gives you a pretty good idea what it looks like
there is a zero left of -1, a zero between 0 and +1, and a zero right of +1
I will leave finding those zeros for you, guess and then guess again
Now You know there is an max around -1 and a min around +1
I would take the derivative to look for where, but you can guess and correct the guess
taking derivative and setting to zero
0 = 3 x^2 -8
x^2 = 8/3
x = +/- 1.63
so we have a max at x = -1.63 , you find y
we have a min at x=+1.63, you find y
Now you can easily make your graph more accurate to answer e and f
Trigonometry concerns triangles. That subject may be part of your book, but is not the subject of your question. Your question is about functions.
0 answers