b/c you didnt answer my questions. shelley" "** Query 3.6.50 sketch f(x) = (x-2) / (x^2-4x+3). **** Describe your graph, including a description of all intercepts, extrema, asymptotes and concavity.
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zeros: x=2
(-i,2)(2,i)
- +
Good reasoning, but the function is not continuous at x = 1 or at x = 3. The same-sign-on-every-interval argument only works for larger intervals on which the function is continuous.
The discontinuities at x = 1 and x = 3 (below) tell us that the function is continuous on (-infinity, 1), (1, 3) and (3, infinity). Each of these intervals must be considered separately.
The function has no zeros on (-inifinity, 1) or on (3, infinity), so the function has a single sign on each interval. Considering negative x values of large magnitude it is clear that on (-infinity, 1) the function will be negative. Considering large positive x values we see that the function is positive on (3, infinity).
The interval (1, 3) has a zero at x = 2, so we need to consider the intervals (1, 2) and (2, 3). It is easy to test each interval, and we find that the value is positive on (1, 2) and negative on (1, 3).
Summary:
We have intervals (-infinity, 1), (1, 2), (2, 3) and (3, infinity) on which the function is respectively negative, positive, negative and positive.
See below to analyze the vertical and horizontal asymptotes, and concavity.
h.a.: y=0 deg num< deg denom
v.a: x^2-4x+3=0 (x-1)(x-3)=0
x=1,3
(-i,1) (1,3) (3,i)
+ - +
The denominator x^2 - 4 x + 3 has zeros at x = 1 and x = 3, as you say.
The numerator is not zero at either of these points, so as you approach these points your denominator approaches zero and your numerator doesn't. The resulting division will approach and unbounded limit, approaching either -infinity or +infinity.
Near x = 1 the numerator x - 2 is negative; near x = 3 the numerator x - 2 is positive.
As you approach x = 1 from the left, x is in the interval (-infinity, 1) so the denominator is positive. As we just saw the numerator is negative, so the limiting value is -infinity.
As you approach x = 1 from the right, x is in the interval (1, 3) so the denominator is negative. As we just saw the numerator is negative, so the limiting value is +infinity.
So at x = 1 we have a vertical asymptote approaching -infinity from the left and +infinity from the right.
Similar reasoning shows that as we approach x = 3 from the left the limiting value is -infinity and from the right the limiting value is infinity. So the description of the x = 3 vertical asymtote is the same as that of the x = 1 vertical asymptote.
As x -> infinity or -infinity the denominator, with its greater power of x, dominates so at both 'ends' we have horizontal asymptotes approaching the x axis.
f(x)=x-2 g(x)=x^2-4x+3
f'(x)=1 g'(x)=2x-4
f'(x)=(x^2+4x+5)/((x^2-4x+3)^2)
I believe this comes out (-x^2 + 4 x - 5) / (x^2 - 4 x + 3)^2; this will still have no critical points.
the numerator is a quad so never zero no crit pts
f''(x)=((x^2+4x+5)(2x-4)^2-(2x+4)(x^2-4x+3)^2)/((x^2-4x+3)^4)
The derivative of f ' would be [ ( -x^3 + 4 x - 5) ' ( x^2 - 4 x + 3)^2 - (x^2 - 4 x + 3) ' ( -x^2 + 4 x - 5) ] / (x^4 - 4 x + 3)^4, which simplifies to f '' = 2(x - 2)(x^2 - 4·x + 7)/(x^2 - 4·x + 3)^3.
The denominator of f '' is zero at x = 1 and x = 3, negative between x = 1 and x = 3 and positive elsewhere. The numerators has two factors; the first is x - 2, which is positive for x > 2 and negative for x < 2. The other factor is x^2 - 4 x + 7, which has a negative discriminant and therefore no zeros. We easily see that this factor is therefore always positive.
So we consider the intervals (-infinity, 1), (1, 2), (2, 3) and (3, infinity). On the first of these intervals the numerator is negative and the denominator positive, so f '' is negative. On the second the numerator remains negative but the denominator is now negative so f '' is positive. Similar straightforward reasoning shows that on the third interval we have pos / neg so f '' is negative; ofer the fourth we have pos / pos and f '' is positive.
The resulting function therefore has the following description:
On (-infinity, 1) the function is negative and concave down, approaching the negative x axis as a horizontal asymptote and the vertical line x = 1 as a vertical asymptote.
On (1, 2) the function is positive and concave up, approaching x = 1 as a vertical asymptote.
At x = 2 the function is zero and its concavity changes from positive to negative.
On (2, 3) the function is negative and concave down, approaching the line x = 3 as a vertical asymptote.
On (3, infinity) the function is positive and concave up, approaching the positive x axis as a horizontal asymptote and the line x = 3 as a vertical asymptote.
f''(x)=((x^2+4x+5)(2x-4)^2-(2x+4))/((x^2-4x+3)^2)
i feel like my math is off here, and im not sure what to do next. i guess you use 2nd deriv to find critical pts and test values...
self critique assessment: 2
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11:28:08
The function has zeros when the numerator is 0, at x = 2; only one x intercept which occurs at x = 2. The numerator is negative on (-inf,2) and positive on (2, inf).
The y intercept is at x = 0; we get (0, -2/3).
Vertical asymptotes when denominator is zero, which occurs at x = 1 and at x = 3. On (-inf,1), (1, 3) and (3, inf) the denominator is respectively positive, negative and positive.
The function is always of one sign between zeros and asymptotes, i.e., on (-inf,1), (1, 2), (2, 3) and (3, inf). Putting together what we know about signs of the numerator and denominator we see that the respective signs on these intervals are negative, positive, negative and positive.
For large pos and neg x values f(x) -> 0 so + and - x axis is an asymptote.
First derivative is
- (x^2 - 4? + 5)/(x^2 - 4? + 3)^2
and 2d derivative is
2?x - 2)?x^2 - 4? + 7)/(x^2 - 4? + 3)^3.
The function is critical when 1st derivative is 0; numerator of 1st derivative is never 0 (quad formula, discriminant is negative) so there are no critical points.
2d derivative:
x^2 - 4 x + 7 is always positive (discriminant negative, expression positive for x=0). Numerator zero only when x = 2.
The denominator is zero at x=3, x=1.
The 2d derivative is therefore of one sign on (-inf,1), on (1, 2), on (2,3) and on (3, inf).
Substitution shows -, +, -, + on these respective intervals and concavity is therefore down, up, down and up.
So the function starts with a horizontal asymptote below the negative x axis, remains negative and concave down to a vertical asymptote at x=1. It then becomes positive with upward concavity, descending from a vertical asymptote at x = 1 to a zero at x = 2, then becoming negative with downward concavity as it approaches a negative asymptote at x = 3. To the right of x = 3 the graph descends from a positive vertical asymptote to a horizontal asymptote at the positive x axis, remaining positive and concave up. **
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****The function is always of one sign between zeros and asymptotes, i.e., on (-inf,1), (1, 2), (2, 3) and (3, inf). Putting together what we know about signs of the numerator and denominator we see that the respective signs on these intervals are negative, positive, negative and positive.I DONT UNDERSTAND THIS STATEMENT. WHICH EQUATION ARE U USING TO TEST HERE?****
****he 2d derivative is therefore of one sign on (-inf,1), on (1, 2), on (2,3) and on (3, inf).
Substitution shows -, +, -, + on these respective intervals and concavity is therefore down, up, down and up. SAME THING DONT GET THIS
self critique assessment: 0
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11:28:24
SOLUTION TO PROBLEM #43:
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not assigned
self critique assessment: 3
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11:29:37
The graph of x y^2 = 4 is not defined for either x = 0 or y = 0. The function has horizontal and vertical asymptotes at the axes.
The graph of x y^2 = 4 is not defined for negative x because y^2 cannot be a negative number.
However for any x value y can be positive or negative. So the first-quadrant graph is also reflected into the fourth quadrant and both are part of the graph of the relation.
You therefore have the graphs of both y = 2 / sqrt(x) and y = -2 / sqrt(x).
The graph of the first-quadrant function will be decreasing since its derivative is negative, and will as you say be asymptotic to the x axis.
The graph of the fourth-quadrant function is increasing and also asymptotic to the x axis. **
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why can't y^2 be negative?
***However for any x value y can be positive or negative. So the first-quadrant graph is also reflected into the fourth quadrant and both are part of the graph of the relation.
You therefore have the graphs of both y = 2 / sqrt(x) and y = -2 / sqrt(x). I DONT FOLLOW THIS
self critique assessment: 0 "