CPhill

We need to find values for "a"  and "b"  that will make the graph continuous

First, let  x  = -2

And solve this for :b"

-2 - 5  =2 (-2) - b

-7 = -4 - b

-3  = -b

b  = 3

Next, let x  = 2  

And solve this for "a"

2 - 5  + a(2)  + 3

-3  = 2a + 3

-6  = 2a

a = -3

So  we have that

                  -3x + 3, if x > 2    

f(x)  =            x - 5,  if  -2 ≤ x ≤ 2

                   2x - 3, if x < -2

Here's the graph from left to right :  https://www.desmos.com/calculator/ibd5jn8m8s   

Answered here :  http://web2.0calc.com/questions/help_63047

Thanks, Gh0sty...!!

Another way  is just to add the whole number and the improper fraction right away

Example

2 +  4/3   =  [ 3*2 + 4] / 3  =  10/3

And of course.....we could write the improper fraction as a mixed number and add

2 +  4/3  =

2 + 1 + 1/3  =

3 + 1/3   =

10 / 3

\(\[h(x) = \left\{ \begin{array}{cl} \lfloor 4x \rfloor & \text{if } x \le \pi, \\ 3-x & \text{if }\pi < x \le 5.2, \\ x^2& \text{if }5.2< x. \end{array}\right.\]\)

Find   \($h(h(\sqrt{2}))$\)

We want to first evaluate   h (√2)....note that   √2 < pi ....so  we want to use the first function

h (√2)   =    floor [ 4(√2) ]   =  floor  [ ≈ 5.657 ]  ....and we want the greatest integer < or  = to 5.657

This  is 5

Now ....we want to evaluate  h(5).....this falls between  pi   and 5.2, so we want to use the second function

So   h(5)  =  3 - 5   = -2

So.....to recap

h ( h(√2) )  = h (5)  =  -2

\(\[\frac{x + 7}{x^2 - x - 2} = \frac{A}{x - 2} + \frac{B}{x + 1}\]\)

\( $x\neq -1$\)

and

\($x\neq 2$\)

This is a partial fractions problem.....we want to solve this

x + 7  =  A(x + 1) + B(x - 2)     simplify

x + 7  = (A + B)x + (A - 2B)     equating coefficients, we have the following system

A + B  = 1    →   B  = 1 - A     (1)

A - 2B  = 7      (2)

Sub (1) into (2)  and we have

A - 2(1 - A)  = 7

3A - 2  = 7

3A  = 9

A  = 3

And B  = 1 - 3  =  -2

So  (A, B)  =  (3 , -2)

x  + 1/x  = -5    (1)  

square  both sides  and we have

x^2 + 2 + (1/x)^2  = 25    →  x^2 + (1/x)^2  = 23    (2)

x^3  +  (1/x)^3      can be factored as a sum of cubes  thusly

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

(x + 1/x) ( x^2 + (1/x)^2  - 1 )    sub (1)  and (2)  into this

(-5) ( 23  -  1)  =

(-5) (22)  = 

-110

There is no x^2y^2 term in the expansion of this.....the sum of the exponents on both variables must = 8

 5^17+5^17+5^17+5^17+5^17

Note that what we really have is

5 *  5^17      or......put another way....

5^1  *  5^17         and remember that  a^m  * a^n  =  a^(m + n)

And we can write

5^1 * 5^17   =  5^( 1 + 17)   =   5^18

Note that the formula for the arc length, S,  is

S = r * theta       where  "theta" is the central angle forming the arc and r is the radius

And let   r  = "r₁"        

So....we have this system

32  = r * theta     →   32/r  = theta    (1)        and  letting r₂ = r + 1.5

34  =  (r + 1.5) * theta         (2)

Sub (1) into (2)   for   "theta"    and we have that

34  = (r + 1.5) (32/r)     multiply through by r

34r =  (r + 1.5) (32)     simplify

34r  = 32r + 48      subtract 32r  from both sides

2r  = 48       divide both sides by 2

r  = 24      =   " r₁ "

And

r + 1.5  =  24 + 1.5  = 25.5  =  "r₂"

LOL!!!....no, but Google Translate does.......

Well, somewhat, anyway.......I have a feeling that something got lost in the translation ....!!!