+0  
 
0
71
11
avatar+25 

For a), I got the answer as -2x^2^-2 +5x^-2.  Is this correct?  And as for part b), I don't know how to do it.  Please help, thanks,

 Nov 3, 2019
 #1
avatar+19857 
-1

First, simplify it to

(x^2-5x-24)/x

x-5 - 24/x

x-5- x^-1 (24)    now take the derivative

1+24 (x^-2)   re-arrange

 

(x^2+24)/x^2

 Nov 3, 2019
 #3
avatar+25 
+1

I thought the derivative of x2-5x-24 would be 2x-5 (the 24 is zero)? 

Umbrella  Nov 3, 2019
 #4
avatar+19857 
0

it is the derivative of   [ x^2-5x-24] / x       Note the red part.....

ElectricPavlov  Nov 3, 2019
 #5
avatar+25 
+1

I turned x into x-1 which differentiates to -x-2 which I then multiplied by the numerators (2x and -5).  Is that wrong?

Umbrella  Nov 3, 2019
 #6
avatar+19857 
0

It's     x-1  (x2-5x-24)     <=====     take the derivative of THIS      ( you see there is a  term  24x-1   you cannot discard this portion )

ElectricPavlov  Nov 3, 2019
 #7
avatar+25 
+1

Ah, I get it now.  Thank you.  So the equation for part b) is y-7x+29=0?

Umbrella  Nov 3, 2019
 #8
avatar+19857 
0

Look at the graph.....what do you think?

 

https://www.desmos.com/calculator/31job5ypvn

ElectricPavlov  Nov 3, 2019
 #9
avatar+25 
+1

Well it's a tangent to the original equation, so I think that it's definitely correct.  

Umbrella  Nov 3, 2019
 #10
avatar+19857 
0

Nailed it.

ElectricPavlov  Nov 3, 2019
 #11
avatar+25 
+1

Thanks a lot for the help. 

Umbrella  Nov 3, 2019
 #2
avatar+19857 
0

The point C will be    x = 2

y = (2+3)(2-8) /2   = -15         2,-15

 

the slope (m) is given by the derivative at x = 2 

    Then just use   y = mx + b   to find your line.....can you do that?

 Nov 3, 2019

4 Online Users

avatar