#### Math 1501 B1C,B2C

I'm Tom Morley

Solutions for quiz 1

• 1. Find all x such that | x^2 - 10 x | <= 16. (<= is less than or equal.).

If | x^2 - 10 x | = 16, then either (x^2 - 10 x ) = 16 or - (x^2 - 10 x ) = 16. The roots of these two quadatic equations are (in order):

```
5-sqrt(41) < 2 < 8 < 5 + sqrt(41)
```
```  At x = 5, | x^2 - 10 x = 0| = | -25| = 25 >  16,
at  x = -3, x^2 - 10 x =  (-3)^2 - 10 *(-3) = 36 >  16
at x = 15,  x^2 - 10 x = 75 > 16
at x = 0 , x^2 - 10 x = 0 < 16
at x = 10 x^2 - 10 x = 0 < 16
```
Therefore, the x such that | x^2 - 10 x | <= 16 are exactly :
```
[5-sqrt(41),2] union [8, 5+ sqrt(41)]
```

• 2. From the definition, compute the derivative of 1/sqrt(2x + 3).
```

( Note --> = limit)

f(x + h ) - f(x)
----------------
h

= 1/sqrt(2(x+h) + 3) - 1/sqrt(2x+3)
---------------------------------
h

= 1/sqrt(2(x+h) + 3) - 1/sqrt(2x+3)  1/sqrt(2(x+h) + 3) + 1/sqrt(2x+3)
---------------------------------   ---------------------------------
h                       1/sqrt(2(x+h) + 3) + 1/sqrt(2x+3)

=        1/(2x+2h+3) - 1/(2x+3)
--------------------------------------
h (1/sqrt(2(x+h) + 3) + 1/sqrt(2x+3))

=   -2h / ((2x+2h+3) (2x+3))
-------------------------------------
h (1/sqrt(2(x+h) + 3) + 1/sqrt(2x+3))

=   -2 / ((2x+2h+3) (2x+3))
-------------------------------------  --->
(1/sqrt(2(x+h) + 3) + 1/sqrt(2x+3))

=   -2 / ((2x+3) (2x+3))
-------------------------------------  --->
(1/sqrt(2x + 3) + 1/sqrt(2x+3))

=       -2 / ((2x+3) ^2)
-----------------------
(2/sqrt(2x + 3)

sqrt(2x + 3)          -1
------------ = ---------------
((2x+3) ^2)      (2x + 3) ^(3/2)

```
• 3. Find all points on the curve x^3 /3 - 2 x ^2 + 7/2 x + 6 such that the tangent line is perpendicular to the line 3 y - 4 = -6x
```The slope of the line is -2, so we are trying to solve

f'(x) = 1/2.

This is

x^2 - 4 x + 7 /2 = 2,  whose solutions are:

x = 1 and x = 3. The points are therefore (x,f(x)) which are:

(1, 47/6)
(3, 15/2)

```