How do you sketch the curve #y=(x+1)/sqrt(5x^2+35)# by finding local maximum, minimum, inflection points, asymptotes, and intercepts?

Answer 1

see below
graph{(x+1)/(sqrt(5x^2+35)) [-10, 10, -5, 5]}

#f(x)=(x+1)/(sqrt(5x^2+35))#
Vertical asymptote is in the point where it isn't defined that is: #sqrt(5x^2+35)!=0# #5x^2+35!=0# #x^2!=-35/5# #x!=sqrt(-35/5)# As you can see this is absolute nonsense. That can't be done. Square root must be #>0#. That means it doesn't have a vertical asymptote and The Domain is #RR#
In order to determine maximum and minimum we need first derivative: #f(x)=(x+1)/(sqrt(5x^2+35))#
#f^'(x)=((x+1)^'(sqrt(5x^2+35))-(x+1)((5x^2+35)^(1/2))^')/(sqrt(5x^2+35))^2#
#f^'(x)=(1*sqrt(5x^2+35)-(x+1)((1*10x))/(2sqrt(5x^2+35)))/(sqrt(5x^2+35))^2#
#f^'(x)=((2(sqrt(5x^2+35))^2-10x(x+1))/(2sqrt(5x^2+35)))/(sqrt(5x^2+35))^2#
#f^'(x)=(2(5x^2+35)-10x(x+1))/(2(sqrt(5x^2+35))^3#
#f^'(x)=(cancel(10x^2)+2*35cancel(-10x^2)-2*5x)/(2(sqrt(5x^2+35))^3#
#f^'(x)=(cancel2(35-5x))/(cancel2(sqrt(5x^2+35))^3#
#f^'(x)=(5(7-x))/(sqrt(5x^2+35))^3#
#sqrt(5x^2+35)# is always positive so we don't care about that. However #(7-x)# can be + and - #x in (-oo,7) hArr f^'(x) >0 =># f goes up #x in (7,oo) hArr f^'(x) <0 =># f goes down
maximum is in x=7#quadf(7)=(7+1)/(sqrt(5*(7)^2+35))=8/(sqrt(280))=4/sqrt70~~0.478#

function doesn't have minimum

horizontal asymptote: #Lim_(xrarr+-oo)f(x)# 1. for #+oo# #Lim_(xrarroo)(cancelx(1+1/x))/(cancelx(sqrt(5+35/(x^2))))=Lim_(xrarroo)(1+1/x)/(sqrt(5+35/(x^2)))=(1+0)/(sqrt(5+0))=1/sqrt5~~0.447#
Intercepts: #if x=0quad=>quad y=1/sqrt35#
#if y=0quad=>quad 0=x+1quad=>quadx=-1#
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Answer 2

To sketch the curve (y = \frac{x+1}{\sqrt{5x^2 + 35}}), we first find the critical points by taking the derivative, set it to zero to find the local extrema, and then find the inflection points and asymptotes. Finally, we plot the curve by incorporating these points.

1. Find the Derivative:
[y' = \frac{d}{dx} \left( \frac{x+1}{\sqrt{5x^2 + 35}} \right)]
[y' = \frac{\sqrt{5x^2 + 35} - (x+1)(\frac{1}{2})(5x^2 + 35)^{-\frac{1}{2}}(10x)}{5x^2 + 35}]
[y' = \frac{(5x^2 + 35) - (x+1)(10x)}{2(5x^2 + 35)^\frac{3}{2}}]
[y' = \frac{5x^2 + 35 - 10x^2 - 10x}{2(5x^2 + 35)^\frac{3}{2}}]
[y' = \frac{-5x^2 - 10x + 35}{2(5x^2 + 35)^\frac{3}{2}}]

2. Find the Critical Points:
Set (y' = 0) to find critical points.
[-5x^2 - 10x + 35 = 0]
[5x^2 + 10x - 35 = 0]
[x^2 + 2x - 7 = 0]
[(x + 3)(x - 1) = 0]
[x = -3, 1]

3. Find the Local Extrema:
Use the second derivative test to find local extrema.
[y'' = \frac{d^2}{dx^2} \left( \frac{x+1}{\sqrt{5x^2 + 35}} \right)]
[y'' = \frac{d}{dx} \left( \frac{-5x^2 - 10x + 35}{2(5x^2 + 35)^\frac{3}{2}} \right)]
[y'' = \frac{-10x - 10}{2(5x^2 + 35)^\frac{3}{2}} - \frac{-5x^2 - 10x + 35}{2(5x^2 + 35)^\frac{5}{2}}(10x)]
[y'' = \frac{-10x - 10}{2(5x^2 + 35)^\frac{3}{2}} + \frac{(5x^2 + 10x - 35)(10x)}{2(5x^2 + 35)^\frac{5}{2}}]

4. Find the Inflection Points:
Set (y'' = 0) to find inflection points.

5. Find Asymptotes and Intercepts:
To find asymptotes and intercepts, analyze the behavior of the function as (x) approaches infinity and negative infinity. Also, find the (y)-intercept by setting (x = 0).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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