Could somebody help !?

[tädif]

I really appreciate, if somebody could help me with these tasks, I am beginner and really need fast help. Also, as my mothertonque is not English, some mathematical terms may be wrong, hope you understand.



1. Function y=ln (x+10/x-8)+80

a)find domain, b)asymptotes equations, c)Extremums, d)grow and decline areas, e)curvature and concava areas, f)point of inflections



2.Function: z=xy+40x, a) find Extremums in condition: x^2+y^2=256

b) control sufficiency, c) designate type, d) calculate the value of function



Thank you!

 

[galactus]

I really appreciate, if somebody could help me with these tasks, I am beginner and really need fast help. Also, as my mothertonque is not English, some mathematical terms may be wrong, hope you understand.

1. Function y=ln (x+10/x-8)+80

a)find domain, b)asymptotes equations, c)Extremums, d)grow and decline areas, e)curvature and concava areas, f)point of inflections

To find the asymptotes, look at the graph. That helps for the other questions as well. Remember, ln(0) is not defined. We can rewrite as:

\(\displaystyle ln(\frac{x+10}{x-8})+80=ln(x+10)-ln(x-8)+80\). What values make ln(0)?. These are NOT in the domain. The domain are those values of x which result in a real solution.

Therefore, the domain would be \(\displaystyle (-\infty,-10)\cap(8,\infty)\)

If you check values such as x=1, there is a complex result so it is not in the domain. See?.

To find the extremums, differentiate. \(\displaystyle f'(x)=\frac{-18}{(x-8)(x+10)}\)

To find inflection points, use the second derivative. \(\displaystyle f''(x)=\frac{36(x+1)}{(x-8)^{2}(x+10)^{2}}\)

The inflection point(s) is where it changes concavity. To find increasing and decreasing. Check the value of the derivatives over an interval.



Set the second derivative to 0 and solve for x. Well, the only value that makes this 0 is \(\displaystyle x+1=0\Rightarrow x=-1\)

There's a start. We can not give a calc I lesson. Look over a calc book or google and you'll find lotsa information regarding this issue.

 

[BigGlenntheHeavy]

\(\displaystyle 10/x-8 \ = \ \frac{10}{x}-8, \ however, \ for \ emphasis \ and \ common \ courtesy, \ \bigg(\frac{10}{x}\bigg)-8 \ = \ \frac{10}{x}-8 \ leaves \ no\)

\(\displaystyle \ doubt \ and \ 10/(x-8) \ = \ \frac{10}{x-8}, \ parentheses \ a \ necessity.\)

\(\displaystyle Now, \ which \ of \ the \ two \ is \ it?\)

\(\displaystyle Note: \ x+10/x-8 \ = \ x+\frac{10}{x}-8, \ (x+10)/x-8 \ = \ \frac{x+10}{x}-8, \ x+10 /(x-8) \ = \ x+\frac{10}{x-8}, \ and\)

\(\displaystyle \ (x+10)/(x-8) \ = \ \frac{x+10}{x-8}.\)

 

[tädif]

Thank you, but how to find asymptotes equations?

 

[galactus]

Thank you, but how to find asymptotes equations?

As I hinted at, the asymptotes occur where at ln(0). What makes ln(x+10)=ln(0)?. What makes ln(x-8)=ln(0)?.