Melody

Let   \(f(x)=\left(\frac37\right)^x\)     be a function that is defined on the domain \([0,\infty)\). Find the range of the function.

f(0)=1     f(infinity)=+0      all the other values in the given domain are between these  so

range (0,1]

The domain is all the possible x values

and

The range is all the range is all the posible y vlues.

You can often get it just by looking at or thinking about the curve.

Sometimes you must look at the eqation and look at what values x or y can, or cannot take.

-----------------

for instance

y-2 =1/x

You cannot divide by 0 so x is not zero

so the domain is all real x where x is not equal to 0      

\(Domain:\:x\in R\;\;where\;\; \:x\ne0\;\;\)

1 divided by anything is not equal to 0 so y-2 is not equal to 0 so    y is not equal to 2

\(Range:\:y\in R\;\;where\;\; \:y\ne2\;\;\)

So x=0 and y=2 are the asymptotes of this hyperbola.

Well the answer is 2 but I do not know how to do it.

It says...

"The diagram below shows the initial position of square DEFG,  an intermediate position, and the final position."

Where is the diagram?

answered.

B<4

B>-10

This is the union of the 2.  It includes the whole number line,

You just need to add the arrows at the ends.  And name the number line B

Suppose the line tangent to the graph of f at x = 2 is y = 4x + 1

This tells you that the point (2,9) is on the f graph

\(f(2)=9\)

The Equation of tangent to the f funtion    at x=2   is    y=4x+1

and suppose the line tangent to the graph of g at x = 2 has slope 3 and passes through (0, -2).

At   f'(2) = 3

And this tangent line, the one at f(2)  passes through (0,-2)

For  this line we are told (0,-2) is a point and the gradient is 3.    We need to find the find the y value for x=2.

\(m=\frac{rise}{run}\\ 3=\frac{y_1--2}{2-0}\\ 6=y_1+2\\ y_1=4\\ g(2)=4 \)

The Equation of tangent to the g funtion    at x=2   is    y=3x-2

a. y = f(x) + g(x) 

f'(2)+g'(2)=4+3=7

f(2)+g(2)= 9+4=13

So the equation of the tangent at x=2 is

y=7x+13

b. y = f(x) - 2g(x)

2g(x) the gradient and the y values will be doubled    so gradient of tangent=6 at (2,8)

f'(2)+2g'(2)=4+6=10

f(2)+g(2)= 9+8 = 17

y=10x-17

c. y = 4f(x)   

tangent of 4f(x)=4(4x+1) at x=2     is   y = 16x+4

I think that is right.

You can try convincing yourself (one way or the other) by drawing some graphs.

It has been fabulous having you here Chris.

You are the heart and soul of this place.   

When the polynomial P(x) = 6x3 + kx2 + x − 2 is divided by x + 2, the remainder is 0. Which of the following is also a factor of P(x)?

A:3x-1

B:3x+1

C:2x-1

D:x-1

If x+2 is a factor then f(-2)=0

so

\(P(2) = 6*(-2)^3 + k*(-2)^2 + (-2)− 2=0\\ 6*-8 + 4k-4=0\\ 4k=52\\ k=13\\ so\\ p(x)=6x^3+13x^2+x-2\)

If 3x-1 is a factor then x=1/3  will be a zero I can check this with the calcuator

6*(1/3)^3+13*(1/3)^2+(1/3)-2 = 0    So  (3x-1) IS a factor

If 3x+1 is a facor then  f(-1/3) = 0

6*(-1/3)^3+13*(-1/3)^2+(-1/3)-2 = -1.1111111111111111    No 3x+1 is not a factor

If 2x-1 is a factor then f(1/2)=0

6*(1/2)^3+13*(1/2)^2+(1/2)-2 = 2.5    No 2x-1 is not a factor

If x-1 is a factor then f(1)=0

6*(1)^3+13*(1)^2+(1)-2 = 18     No that isn't a factor either     

I have determined this without any algebraic division.   

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