Question

How many points with integral coordinates are on the curve y=(x-3)/(x^2-x)?

Answer 1

Answer:(−1,−2) and (3,0).

Step-by-step explanation:

We want to determine how many points are there with integral coordinates on the curve y=x−3/x(x−1).  

Let us see how this curve behaves.

y=0 for x=3.

dy/dx=−x−6x+4/(x^{2})(x-1)x^{2}).

At the critical points, the first derivative is zero ⇒x2−6x+4=0.

⇒x=3±5–√⇒x≈0.764 or x≈5.236.

It can be verified by the second derivative test that the function has a local minimum at x≈0.764 and a local maximum at x≈5.236.

Since we have x(x−1) in the denominator, x cannot take the values 0 and 1.

Further, it can be seen that for x>3 and for 0<x<1, the function is positive and for x<0 and for 1<x<3, the function is negative.

As x tends to 0+ or 1−,y tends to ∞ and as x tends to 0− or 1+,y tends to −∞.

⇒ The domain of the function is (−∞,0)∪(0,1)∪(1,∞).

Interval (−∞,0)

In this interval, the first derivative, −x2−6x+4/(x^{2})(x-1)x^{2}), is always negative.

⇒ In this interval, the function is monotonically decreasing.  

f(−2)=−56<1.

⇒ The function does not have integral coordinates for x≤−2.

f(−1)=−2.  

⇒ The function has an integral coordinate at (−1,−2).

Interval (0,1)

There are no integer values of the variable in the interval (0,1).

Interval (1,∞)

f(2)=−12.

f(3)=0.

⇒ The function has an integral coordinate at (3,0).

f(3+$\sqrt{5}$5)=$\sqrt{5}$5/11+5$\sqrt{5}$5 < 1.

For x>3, the function takes only positive values, y tends to 0 as x tends to ∞ and the maximum of this function in this interval is $\sqrt{5}$5/11+5$\sqrt{5}$5 < 1.

⇒ The function does not have integral coordinates for x>3.

We therefore conclude that the function has only two integral coordinates, which are (−1,−2) and (3,0).

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Answer 2

Answer:

(-1,-2) and (3,0)

Step-by-step explanation:

We want to determine how many points are there with integral coordinates on the curve y=x−3/x(x−1).  

Let us see how this curve behaves.

y=0 for x=3.

dy/dx=−x−6x+4/(x^{2})(x-1)x^{2}).

At the critical points, the first derivative is zero ⇒x2−6x+4=0.

⇒x=3±5–√⇒x≈0.764 or x≈5.236.

It can be verified by the second derivative test that the function has a local minimum at x≈0.764 and a local maximum at x≈5.236.

Since we have x(x−1) in the denominator, x cannot take the values 0 and 1.

Further, it can be seen that for x>3 and for 0<x<1, the function is positive and for x<0 and for 1<x<3, the function is negative.

As x tends to 0+ or 1−,y tends to ∞ and as x tends to 0− or 1+,y tends to −∞.

⇒ The domain of the function is (−∞,0)∪(0,1)∪(1,∞).

Interval (−∞,0)

In this interval, the first derivative, −x2−6x+4/(x^{2})(x-1)x^{2}), is always negative.

⇒ In this interval, the function is monotonically decreasing.  

f(−2)=−56<1.

⇒ The function does not have integral coordinates for x≤−2.

f(−1)=−2.  

⇒ The function has an integral coordinate at (−1,−2).

Interval (0,1)

There are no integer values of the variable in the interval (0,1).

Interval (1,∞)

f(2)=−12.

f(3)=0.

⇒ The function has an integral coordinate at (3,0).

f(3+5)=5/11+55 < 1.

For x>3, the function takes only positive values, y tends to 0 as x tends to ∞ and the maximum of this function in this interval is 5/11+55 < 1.

⇒ The function does not have integral coordinates for x>3.

We therefore conclude that the function has only two integral coordinates, which are (−1,−2) and (3,0).

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