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Absolutely! Otherwise, the expression is not completely simplified. 

I multiply 3 with the trinomial, then expand the two other binomials?

If \(f(b)=4\), then this means that there is a specific input for this function that outputs 4. On a graph, the x-values represent the inputs of the function and the y-values represent the output. In this case, \(f(2)=4\), which is evident because of the given graph. 

If \(f(2)=4\), then \(b=2\), so \(f(a)=2\). Using the same reasoning as before, \(f(0)=2\), so \(a=0\). 

\(a-b=0-2=-2\)

Hello Guest, 

I have not scrutinized your work thoroughly yet, but I think you have done everything correctly thus far. In order to make progress here, you must simplify the numerator \(3(2x^2+9x-5)+(4x^2-1)(x-5)\). Do you know how to proceed from here?

f(-x)  reflects the  graph of f(x) across the y axis.....the  x coordinate becomes  - 2  and the y coordinate remains the same

So....the point (-2, 9)   is on the graph  of  f(-x)   and the sum of these coordinates is  7

2h / (h^2 - 9)  +  h / (h^2 + 6h + 9)   - 3/ ( h - 3)          factor the denominators

2h / [ ( h + 3) ( h - 3) ]  +  h / [ ( h + 3) (h + 3) ]  - 3 / (h - 3)

The common denominator  is   ( h + 3) ( h + 3) ( h - 3)  ....so we have

2h (h + 3)   + h( h -3)   - 3 (h + 3) ( h + 3)

________________________________       

 (h + 3) ( h + 3)  ( h - 3)

Simplify  and note that  x cannot  be either 3  or  -3  because both values make the denominator  = 0

So we have

2h^2 + 6h + h^2 - 3h  - 3 ( h^2 + 6h + 9)

______________________________

( h + 3) ( h + 3) ( h - 3)

2h^2 + 6h + h^2 - 3h  - 3h^2 - 18h - 27

_______________________________

(h + 3) ( h + 3) (h - 3)

         -15h - 27

___________________

(h + 3) ( h + 3) ( h - 3)

   -3(5h + 9)

___________________

(h + 3) ( h + 3) ( h - 3)

The answer is 40.

We know that the answer must be a multiple of 5 between 30 and 50. But it also said it is a multiple of 8, so it is 40.

f(x) = 7x + 5

g(x) = x - 1

f(g(x))  = 7 (x - 1) + 5  =  7x - 7 + 5 =  7x - 2  =  h(x)

So...for h(x)  write  y

y = 7x  - 2     add 2 to both sides

y + 2 =  7x      dicvide both sides by 7

(y + 2)  / 7   = x       "swap" x and y

(x + 2)/ 7   = y      for y , write h^-1(x)

So

h^-1 ( x)  = (x + 2) / 7

\(f(x)=1-\sqrt{1-x^2} \)

\(-1\le x\le 1 \)

We need to find the inverse of the  given function....for f(x), we can write, y

y = 1 - √  [1 - x^2]      rearrange as

√  [1 - x^2]   = 1 - y      square both sides

1 - x^2  = (1 - y)^2

1 - x^2   =  y^2 - 2y + 1    subtract 1 from both sides

-x^2  = y^2  - 2y        mutiply through  by -1

x^2  = 2y  - y^2        take both roots

x  = ±√ [ 2y - y^2 ]      "swap"  x  and  y

y  = ±√ [2x - x^2 ]

We only need the top half of this graph   ⇒   y  = √ [2x - x^2 ]

Here is the graph  of  both functions  along with the graphs of  y  = x   and y  = 1

https://www.desmos.com/calculator/ozvzaxlwxe

The area enclosed by both functions will be twice the area of the area between  y = x  and  y  = 1 - √  [1 - x^2]

Note that the area  bounded  by the y axis from 0 to 1 , the line y = 1 from 0 to 1   and the line  y  =  x   will from a triangle with a base  and height of 1....so...it's area =  (1/2) base * height  = (1/2)(1)(1)  = 1/2  units^2

And note that the area formed  y axis from 0 to 1, the line y  =1 from 0 to 1  and the function  y = 1 - √  [1 - x^2]    will be the area of a quarter circle with a radius of  1  =

(1/4) pi * (1)^2   =  pi /4  units^2

So....this area  less the area of the triangle will   =  1/2  of the area we are seeking

So....the total area  =  2 [ pi/4  -1/2] =   [ pi/2  - 1] units^2 ≈   0.571  units^2

Here's one more way :

4(x^2 - 9)^2  - ( x + 3)^2        factors as

4 [ (x + 3)(x - 3)] ^2  - (x + 3)^2 =

[(x + 3}^2  [ 4 ( x + 3)^2 ] - 1 ]

And x^2 + 6x + 9   factors as  ( x + 3)(x + 3)  = (x + 3)^2

So we have

[(x + 3}^2  [ 4 ( x - 3)^2 ] - 1 ]

_______________________         we can "cancel the ( x + 3)^2  terms  and we are left with

   (x + 3)^2

4( x - 3)^2 - 1  =

4(x^2 - 6x + 9)  - 1 =

4x^2 - 24x + 36 - 1 =

4x^2 - 24x + 35       

We need two factors that multiply to 4  and 35  with the two factors multiplying to 35 both being negative

We might try  4,1  and  -5 , -7

(4x - 5) ( x - 7) =  4x^2 -5x - 28x + 35  = 4x^2 - 33x + 35....no good

Next...we might guess 2,2  and  -5, -7

(2x - 5) (2x - 7)   = 4x^2 - 10x - 14x + 35  =  4x^2 - 24x + 35   which is what we  need

Also....x  cannot equal -3  because this would make  the origianl denominator   = 0 

1 /  (  l x^2 + 3x  - 4 l + l x^2 + 9x + 20 l )

Note that we  cannot have   x^2 + 3x - 4    and x^2 + 9x + 20    both equaling 0 at the same time

So.....factoring   x^2 + 3x  -  4  and setting it to 0, we get that

(x + 4) ( x - 1)    = 0

Setting each factor to  0 and solving for x, we get that   x = -4   or  x =  1

Similarly....factoring x^2 + 9x + 20   and setting this to  0, we get that

(x + 4) ( x + 5)    =  0

Setting each factor to 0 and solving for  x, we get that x = -4 or  x  = -5

So...it appears the the  real  value that makes both parts of the denominator  = 0  is when x  = -4

See the graph, here : https://www.desmos.com/calculator/ssgeajyewr

4. The graph of $y=f(x)$ is shown below in red. Given that $f$ is invertible, find $f^{-1}(4)$.

5. What are the coordinates of the points where the graphs of $f(x)=x^3-x^2+x+1$ and $g(x)=x^3+x^2+x-1$ intersect? Give your answer as a list of points separated by semicolons, with the points ordered such that their $x$-coordinates are in increasing order. (So "(1,-3); (2,3); (5,-7)" - without the quotes - is a valid answer format.)

Heureka,

no worries, I can see why not....thank you very much!!

Heureka,

no worries, I can see why not....thank you very much!!

If you look closely at the denominator, if x = 2 then you have: x - 2 =2 - 2 =0!!. You should know by now that you cannot divide by zero, because it is "undefined".

thanx guest,

but your answer reads difficultly..

thanx guest,

but your answer reads difficultly..

Heureka,

thank you very much, although I fail to understand something:

why can't x be equal to 2?

Heureka,

thank you very much, although I fail to understand something:

why can't x be equal to 2?

Solve for x:
(x + 2)/(x^2 - 2 x) - 2/(x - 2) = -1

Bring (x + 2)/(x^2 - 2 x) - 2/(x - 2) together using the common denominator x:
-1/x = -1

Take the reciprocal of both sides:
-x = -1

Multiply both sides by -1:

x = 1

solve for X
\dfrac{x+2}{x^2-2x} - \dfrac{2}{x-2}=-1

\(\dfrac{x+2}{x^2-2x} - \dfrac{2}{x-2}=-1\)

\(\begin{array}{|rcll|} \hline \dfrac{x+2}{x^2-2x} - \dfrac{2}{x-2} &=& -1 \quad | \quad x-2 \ne 0 \Rightarrow x \ne 2 \\\\ && \quad | \quad x^2-2x \ne 0 \Rightarrow x \ne 2 \\\\ \dfrac{(x+2)(x-2)-2(x^2-2x)}{(x^2-2x)(x-2)} &=& -1 \\\\ (x+2)(x-2)-2(x^2-2x) &=& -(x^2-2x)(x-2) \\ x^2-4-2x^2+4x &=& -(x^2-2x)(x-2) \\ -x^2+4x-4 &=& -(x^2-2x)(x-2) \quad | \quad \cdot (-1) \\ x^2-4x+4 &=& (x^2-2x)(x-2) \\ (x-2)(x-2) &=& (x^2-2x)(x-2) \quad | \quad : (x-2) \\ x-2 &=& x^2-2x \\ x^2-2x &=& x-2 \\ x^2-3x+2 &=& 0 \\ (x-1)(x-2) &=& 0 \\\\ x = 1 && x \ne 2 \\ \hline \end{array}\)

wow Heureka,

thanx a million!

factorise

4x^2-4xy+y^2-9

\(4x^2-4xy+y^2-9\)

\(\begin{array}{|rcll|} \hline && 4x^2-4xy+y^2-9 \\ &=& (4x^2-4xy+y^2) -9 \\ &=& (2x-y)(2x-y) -9 \\ &=& (2x-y)^2 -9 \\ &=& (2x-y)^2 - 3^2 \\ &=& [(2x-y) - 3][(2x-y) + 3] \\ &=& (2x-y - 3)(2x-y + 3) \\ \hline \end{array}\)

Probability worded problem

Jane has three bags of lollies. In bag 1 there are three mints and three toffees,
in bag 2 there are three mints and two toffees and in bag 3 there are two mints and one toffee.
Jane selects a bag at random and then selects a lolly at random.

c the probability hat jane chose bag 1, given that she selects a mint.

\(\text{P(bag 1)$ = \dfrac13 $} \\ \text{P(mint/bag 1)$ = \dfrac36$} \\ \text{P(mint form bag 1)$ =$ P(bag 1)P(mint/bag 1)$ = \left(\dfrac13\right)\left(\dfrac36\right) = \dfrac16 $}\)

1. The graph of $y=f(x)$ is shown in red below. For how many values of $c$ is $f(c) = -1$?

     4 values