Expert: Scotto - 10/14/2009

Question
My problems (set 3):
9. Find the slope-intercept form of the line satisfying the given condition. The line has an angle of inclination of  =(π/6) and a y-intercept of (-3).
12. What is the solution set of 4(x-2)<(8x-4)<2(x-1)?
14. Find x and y if (4,-5) is the midpoint of the line segment joining (-3,2) and (x,y).
20. To the nearest degree, find the angle of inclination of y+(2x)+5=0.

My problems (set4):
[Ex.6] Find the limit of (3-Radical(x+5))/(x-4) as x approaches 4.
1. Find the limit of (x³-1)/(x-1)²
2. Consider the function (this is a piecewise function) f(x)=(a+bx), if x>2 or 3, if x=2 or if (b-ax²), if x<2. Determine the values of constants (a) and (b) so that limit of f(x) as x approaches 2 exists.

Answer
Set 3
9. Find the slope-intercept form of the line satisfying the given condition. The line has an angle of inclination of  =(π/6) and a y-intercept of (-3).

The slope is the same as the tan(A), where A is the angle.
The tan(π/6) is known to be √3/3.
The standard line is y = mx + b, where m is the slope and b is the y intercept.
The line is then y = √3x/3 - 3.


12. What is the solution set of 4(x-2)<(8x-4)<2(x-1)?

Take 4(x-2) < 8x - 4 and 8x - 4 < 2(x-1) separately.
For the 1st one, multiply out and get 4x - 8 < 8x - 4.
Add 4 - 4x to both sides, giving -4 < 4x, or, dividing by 4, the result is -1 < x,

For the 2nd one, 8x - 4 < 2(x-1), multiply out getting 8x - 4 < 2x - 2.
Add 4 - 2x to both sides, giving 6x < 2, or x < 2/6 = 1/3, so x < 1/3.

Combining both gives -1 < x < 1/3.


14. Find x and y if (4,-5) is the midpoint of the line segment joining (-3,2) and (x,y).
For (4,-5) to be the midpoint of (-3,2) and (x,y), we need to have
(-3+x)/2 = 4 and (2+y)/2 = -5.

For (-3+x)/2 = 4, multiply both sides of the 1st equation by 2, then add 3 to both sides.
(-3+x)/2 = 4 => -3 + x = 8 => x = 11.

For (2+y)/2 = -5, multiply both sides by 2 and then subtract 2 from both sides.
That is, (2+y)/2 = -5 => 2 + y = -10 => y = -12.


Another way to look at it is to get from -3 to 4, 7 is added, so add 7 more to 4 giving 11 for x.
To get from 2 to -5, 7 needs to be subtracted, so subtract 7 more and get -12.



20. To the nearest degree, find the angle of inclination of y+(2x)+5=0.

Change y + 2x + 5 = 0 to y = -2x - 5 by subtracting 2x+5 from both sides.
This says that the slope is -2.  It is known that the slope is the tan(A),
where A is the angle of inclination.  This says that A is 63.435°.
Rounded to the nearest degree gives 63°.


Set 4
[Ex.6] Find the limit of (3-√(x+5))/(x-4) as x approaches 4.

As x->0, the top and bottom both go to 0.
We can then apply the theorem that states if we have f(x)/g(x) going to 0/0 as x->x0,
then f(x) and g(x) can both be differentiated to get the result.

The derivative f'(x) = -1/(2√(x+5)), and f'(4) = -1/(2√9) = -1/6.  The derivative g'(x) = 1.
It can then be seen that f'(x)/g'(x) = (1/6)/1 = 1/6.


1. Find the limit of (x³-1)/(x-1)²
I will assume that we are finding the limit as x->1.
The top factors into (x-1)(x²+x+1).

It can be seen that there is an x-1 on the top and bottom of that fraction,
so we get (x²+x+1)/(x-1).

When x=1 is put in, the result is 3/0, so the answer is ∞.  In other words, its not defined.


2. Consider the function (this is a piecewise function) f(x)=(a+bx), if x>2 or 3, if x=2 or if (b-ax²), if x<2. Determine the values of constants (a) and (b) so that limit of f(x) as x approaches 2 exists.

This says that when 2 is put in, both equations need to be 3 in order to be continuous.
Thus, a + 2b = 3 and b - 4a = 3.

Multiplying the 2nd equation by 2 gives 2b - 8a = 6.
Subtracting off a + 2b = 3 from that gives –9a = 3, so a = -1/3.
Since b - 4a = 3, put in a = -1/3 and get b + 4/3 = 9/3, so b = 5/3.

Checking the work gives a + 2b = -1/3 + 10/3 = 9/3 =3, so that checks.
Also, b - 4a = 5/3 + 4/3 = 9/3 = 3, and so that also checks.

Yeah, that was a long one, but I didn't have that much else to do this time.