Quick calculus question - Need help ASAP?

Question

I have to do a tiny bit of calc - it has been 2 yrs sincce I have done it - can someone pleae help (and show steps)?
Thanks in advance!
I will choose a best answer!

I need to graph the following:
f(x)=4x-8/x^2

I tried to get this to factor out (if that is what I am supposed to do)- perhaps I am looking at the problem incorrectly. I treid using my TI-83 - it give me an error.

PLEASE HELP!!

Answer 1

For x = -2, f(-2) = -8 - 4 = -12
For x = 1, f(1) = 4 - 8 = -4
For x = 2, f(2) = 8 - 8/4 = 6
For x = 3, f(3) = 12 - 8/9 = eleven and one-ninth
For x = 4, f(4) = 16 - 8/16 = fifteen and one-half

As x approaches 0 from either side, the value of the function approaches negative ∞.

As x gets very large, the graph approaches y = 4x from below as an asymptote. You can see that even by the time you get to x = 4, the graph gets pretty close to y = 4x.

Graph those values. You don't need calculus.

Nice explanation by Wal. Openpsych has the value of f(-2) wrong.

Let's take the first derivative of this and see where the curve turns.

f'(x) = 4 + 16 / x^3
0 = 4x^3 + 16
x^3 = -4
x = -cube root(4)

This is a local maximum. For x less than -cube root(4), the curve approaches x = 4y from below. For x greater than -cube root(4), the curve approaches negative ∞.

Answer 2

f(x)=4x-8/x^2

Firstly 8/x^2 is always a positive quantity so there is always something being subtracted from 4x

Secondly 8/x^2 is undefined when x = 0 (cannot divide by 0) so there is a vertical asymptote at x = 0 and it points down as zero is approached from both sides as 8/x^2 is positive

Thirdly as x gets very big in both the positive and negative directions 8/x^2 gets very small so the amount being subtracted from 4x diminishes as x →±∞ so the curve is asymptotic to y = 4x as x →±∞ and always below it.

That should get you started.

Answer 3

Fx=4x-8/x^2
First of all drawing a graph is not calculus
Any way we will attempt this question.
x=0 ,Fx=minus infinity
x=1, Fx=-4
x=4,Fx=15.5
x=1/2, Fx=-30
x=-1,Fx=-12
x=-2,Fx=-6
Now you can plot these values on a graph paper. The graph you will find is asymptotic to y axis for both positive and negative values of x

Answer 4

Problem 16

Let f(x) = 2x - 4 and g(x) = x2 + x

Find f(3)
Solution

f(3) = 2(3) - 4 = 6 - 4 = 2


Find 2f(x) - g(x)
Solution

2f(x) - g(x) = 2(2x - 4) - (x2 + x) Multiply through

= 4x - 8 - x2 - x Reorder and combine like terms

= -x2 + 3x - 8


(f(x))(g(x))
Solution

(f(x))(g(x)) = (2x - 4)(x2 + x) FOIL

= 2x3 + 2x2 -4x2 - 4x Combine like terms

= 2x3 - 2x2 -4x


f(x + h)
Solution

We plug in x + h for x to get

f(x + h) = 2(x + h) - 4

= 2x + 2h - 4

Answer 5

To do this graph you need to find

1. Where it cuts the x and y axis
2. The co'ords and nature of the stationary points

good luck!

Answer 6

Plug this in:
y = 4x - 8x^-2