determine the slope of the graph at the indicated domain value g(x)=x^2+3/x^2-x ;x=2show process plz?

Answer 1

In order to find the instantaneous slope of a curve at a given point one needs to find the derivative of the function and evaluate it at that point.

I assume you know how to find the derivative of a polynomial function.
g(x)=x^2+3/x^2-x
g'(x) = 2x + -6/x^3 - 1

To find the slope of the line at point x = 0, find g'(2).
I find the slope to be 2.25

Answer 2

first we take the derivative. you can do this term by term because there are no products. so we get
g'(x) = 2*x - (3*-2)*x^-3 - 1
= 2*x - 6*x^-3 -1

evaluate at x = 2
g'(2) = 2*2 - 6*(1/8) - 1 = 4 - .75 -1 = 2.25

Answer 3

you need to find g'(x) ie : g'(x) = 2x - 6/x^3 - 1,

then the slope g'(2) = 3 - 6/8 = 3 - 3/4 = 9/4.

Answer 4

The answer is -5/2

Let n=numerator
d=denominator

g'=(n'*d-d'*n)/d^2

n'=2x
d'=2x-1

g'=[2x*(x^2-x)-(2x-1)*(x^2+2)]/(x^2-x)^2)

slope for x=2 is
(4*2-3*6)/4=(8-18)/4=-5/2

Answer 5

from my expriance of being an outstanding POET i will say that it really is okay 6/10 hehheheheheheheh i'm joking it really is tremendous yet has lack of rhyme bas @very last pharoh u recognize what make someone a poet akeed ~~~~~~(i dont recognize if this rule applies to all poets our maximum of them lolz)

Answer 6

i used a calculator and the answer i got was 2.25

Answer 7

-2.5