Sketch The Graph Of F Label Relative Extrema Point Of Inflection And The Asymptote

Sketch the graph of the function. Indicate the asymptotes, local extrema, and points of inflection?

x=0, x=1 are vertical asymptotes
y=x^(-1) - (x-1)^(-1)
y' =(-1)x^(-2) - (-1)(x-1)^(-2)
y'= -1/x^2 + 1/(x-1)^2
y'= - [ (x^2-2x+1)-x^2] / x^2(x-1)^2
y' = -[ -2x+1] / x^2(x-1)^2 =0
2x-1=0
x=1/2 is the critical point
y'' = (-1)(-2)x^(-3) - (-1)(-2)(x-1)^(-3)
y'' = 2/x^3 - 2/(x-1)^3
when x=1/2, y'' =32 > 0, showing that y has a minimum at x=0.5
To find the point(s) of inflection, solve y''=0
2(x-1)^3-2x^3 / x^3(x-1)^3 =0
2(x-1)^3-2x^3=0
2(x^3-3x^2+3x-1) -2x^3=0
-6x^2-6x-2=0
x^2+6x-2=0
This equation is of form ax^2+bx+c
a = 1 b = 6 c = 2
x=[-b+/-sqrt(b^2-4ac)]/2a]
x=[-6 +/-sqrt(6^2-4(1)(2)]/(2)(1)
discriminant is b^2-4ac =28
x=[-6 +√(28)] / (2)(1)
x=[-6 -√(28)] / (2)(1)
x=[-6+5.291502622129181] / 2
x=[-6-5.291502622129181] / 2
The roots are -0.3542 and -5.6458
Points of inflection -0.3542 and -5.6458

How do you find the relative extrema and asymptotes?

plug that into a graphing calculator. Now, the highest point you see in your calculator screen is the "local maxima" same with the lowest visible point, called the "local minma."
Now do you see the lines being straight instead of curved? Those are asymptotes. If it a horizontal line, find the Y-value that it is on, and that is the asymptote for that line. Just, "y=5." And its the same with every other asymptote. just "x= " and "y= "

Find any intercepts, relative extrema, points of inflection, and asymptotes. (If an answer does not exist, en?

Did you mean: y =( x^2)/(x^2 + 243 )
On which part are you having trouble?

intercepts: plug in zero for x ; then plug in zero for y:
(x, y) = (0,0) is the only intercept

relative minimum : find the first derivative: set =0
A fraction is zero when the numerator is zero.
Y' =[ (x^2+ 243)(2x) -(x^2)(2x)]/(x^2+243)^2
= (486x)/(x^2+243)^2

486x= 0
X= 0
(x, y) = (0,0)

relative maximum (none)
(x, y) = ________

points of inflection : set the second derivative = 0
[(X^2+243)^2* 486 - (486x)* 2(x^2+ 243)(2x)] / (x^2+ 243)^4= 0
Set the numerator = 0

X= +/-9
Plug in to the original function to find y:
(x, y) =(-9, 0.25) (smaller x-value)

(x, y) =(9, 0.25) larger x-value)

Find the equation of the asymptote_____________
There is no vertical asymptote, since the denominator cannot= 0.
For the horizontal, simplify the fraction of the largest degrees terms in the numerator and denominator: (x^2)/(x^2)
Y= 1

I hope this helps!

Analyze and sketch the graph of the function. label any intercepts, relative extrema, points of inflection,?

"label any intercepts" can be done without calculus: It means find where it cuts the y axis:
f(0) =(0 - 0 + 9)/(0 - 3) = -3
so the graph cuts the vertical axis at (0, -3)
For the horizontal axis, solve f(x) = 0.
The only way a fraction can be 0 is for the numerator to be zero, so solve
x^2 - 3x + 9 = 0
No real roots, so it doesn't cut the horizontal axis.
Although the question didn't ask for asymptotes, you need to know where they are to draw the graph sensibly. The denominator is 0 when x = 3, so x = 3 is a vertical asymptote.

If we write the top as x(x-3) + 9, we can divide these two terms separately by x-3 and get
f(x) = x + 9/(x - 3)
and as x --> infinity, 9/(x-3) --> 0 , and so
f(x) --> x
Therefore the line y = x is also an asymptote.

To find extrema, you need to solve f '(x) = 0.
i.e. x(x-6) = 0 [I'm trusting your differentiation -- haven't checked it.]
Therefore there are stationary points at (0, -3) and where x = 6,
i.e. at (6, 9)
[f(6) = (36 - 18 + 9)/(6-3)
..... =27/3

Now I can guess that (0, -3) is a maximum turning point, and (6, 9) is a minimum turning point, and there are no inflexions, we have two branches of a hyperbola, one in the top V region formed by y = x and x = 3, the other one opposite to it in the lower region, the inverted V. But let's check formally:

Using f(x) = x + 9/(x-3)
we get f '(x) = 1 - 9/(x-3)^2
This is equal to the expression you got, but is much easier to differentiate again:
f "(x) = 18/(x-3)^3, by using the "function of a function" rule on
1 - 9(x-3)^(-2)

Now f "(0) = 18/(-27) which is negative, therefore (0, -3) is a maximum turning point.
f "(6) = 18/27, positive, so (6, 9) is a minimum turning point.
Those are the two "relative extrema".
18/(x-3)^3 cannot be 0, therefore there is no point of inflexion

Help finding relative & absolute extrema & points of inflection for f(x) = xe^x?

(a) considering the fact that x^2 + a million >= a million for all quite x, the denominator of the rational function f(x) is nonzero for all quite values of x. So the section is all quite numbers. (b) by skill of way of actuality the degree of the numerator, x, is decrease than the degree of the denominator, x^2 + a million, lim x-->infinity f(x) = 0 and lim x-->-infinity f(x) = 0. The horizontal asymptote is y = 0. (c) considering the fact that we already pay interest to the denominator is nonzero for all quite x, there are no further any vertical asymptotes. (d) Use the quotient rule to discover f '(x): f '(x) = [(x^2 + a million)*(d/dx x) - x d/dx (x^2 + a million)]/(x^2 + a million)^2 = [(x^2 + a million)(a million) - x(2x)]/(x^2 + a million)^2 = (a million - x^2)/(x^2 + a million)^2. be conscious that f '(x) is 0 at x = +-a million, efficient mutually as -a million < x < a million, and unfavourable mutually as the two x < -a million or x > a million. So f(x) decreases on (-infinity, -a million) U (a million, infinity) and will strengthen on (-a million, a million). (e) Use the quotient rule lower back to discover f ''(x): f ''(x) = [(x^2 + a million)^2*(d/dx (a million-x^2)) - (a million-x^2) d/dx (x^2 + a million)^2]/(x^2 + a million)^4 = [(x^2 + a million)^2*(-2x) - (a million-x^2)(2)(2x)(x^2 + a million)]/(x^2 + a million)^4 = [(x^2 + a million)(-2x) - (a million-x^2)(2)(2x)]/(x^2 + a million)^3 = [-2x^3 - 2x - (4x - 4x^3)]/(x^2 + a million)^3 = (2x^3 - 6x)/(x^2 + a million)^3 = 2x(x^2 - 3)/(x^2 + a million)^3. be conscious that f ''(x) is 0 at x = 0, +-sqrt(3), efficient mutually as the two -sqrt(3) < x < 0 or x > sqrt(3), and unfavourable mutually as x < -sqrt(3) or 0 < x < sqrt(3). So f(x) is concave down on (-infinity, -sqrt(3)) U (0, sqrt(3)), and concave up on (-sqrt(3), 0) U (sqrt(3), infinity). (f) From the top results of (d), we see that f(x) has a relative minimum at -a million and a relative optimal at a million. (g) From the top results of (e), the concavity variations in any admire 3 x values 0, +-sqrt(3). So components of inflection are at x = 0, +-sqrt(3). (h) regrettably, i visit't draw a graph on yahoo suggestions. optimistically, the recommendations indoors the previous components, alongside with the actuality that f(0) = 0 and f(x) is an outstanding function (its graph is beginning place symmetric), might desire to help you to comic strip the graph of f(x). Lord bless you on the instantaneous!

Analyze and sketch a graph of the function. Find any intercepts, relative extrema, points of inflection, and a?

x-intercept means y = 0
0 = (x^2 + 9)/(x^2 - 49)
0 = (x^2 + 9)
-9 = x^2 ---> no real solutions so DNE

relative minimum ---> DNE

y ' = [ (x^2 - 49) * 2x - (x^2 + 9) * 2x ] / [ (x^2 - 49)^2 ]
0 = [ 2x^3 - 98x - 2x^3 - 18x ] / [ (x^2 - 49)^2 ]
0 = [ 2x^3 - 98x - 2x^3 - 18x ] / [ (x^2 - 49)^2 ] ---> so far x≠ -7 & 7
0 = - 116x
0 = x ---> critical point

check this critical point whether it is a max or min into the first derived equation.
+++-------
_______
. .. 0. . .

f(0) = ( 0^2 + 9 ) / (0^2 - 49) = -9/49
relative maximum (x , y) = ( 0 , -9/49 )

point of inflection = DNE

Vertical asymptotes
x = -7
x= 7

Horizontal asymptote:
lim (x^2 + 9)/(x^2 - 49) ---> large in charge
x---> ∞

lim (x^2/x^2)
x---> ∞

lim 1 = 1
x---> ∞
y = 1 <--- horizontal asymptote

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