Question 3Question 3a) Given the function: f(x) = 3(x-2)^2 + 2, determine the derivative of the function and find the equation of the line tangent to f(x) when x is 4.

b) The points of intersection between the lines tangent to g(x) when x=4 and the line tangent to f(x) when x=5 are the centers of three circles with a radius of 5. Determine the equations of the circles where g(x)= squareroot[x].

Part APart Aa) Given the function: f(x) = 3(x-2)^2 + 2, determine the derivative of the function and find the equation of the line tangent to f(x) when x is 4.

First of all, what is a derivative?

•In mathematics, the derivative of a function is the graph of the function's slope or its rate of change, at any given point. •In general, a derivative is something has been derived or come from something else

The process of finding a derivative is called differentiation.

This question involves fundamental knowledge on derivatives.

Intro.Intro.a) Given the function: f(x) = 3(x-2)^2 + 2, determine the derivative of the function and find the equation of the line tangent to f(x) when x is 4.

The derivative of any constant is zero. So the derivative of 5, 10, 11 is “zero.” To understand this, graph y=5, y=10, and y=11. They all have no slope, so the derivative is zero.

The derivative of any variable to the exponent “1” is “1”. To understand this, graph y=x. X is the variable. You should see through rise over run that the slope is “1.” Y=x is also just another form of y=mx+b where m=1.

Common Notation

This is a common form of Leibniz’s Notation.

This is Lagrange’s notation, also known as prime notation.

Intro.Intro.It is hard to explain this stuff since I’m fairly new to

this stuff, so I won’t go into very much detail.

Power RuleThe derivative of a power of x is equal to the

product of the exponent times x with an exponent reduced by 1

For example: Let’s find the derivative of x^6

You can see that n=6. The derivative of x^6 is 6x^5.

Now let’s proceed to my question.

Part A ContinuedPart A Continueda) Given the function: f(x) = 3(x-2)^2 + 2, determine the derivative of the function and find the equation of the line tangent to f(x) when x is 4.

First, you put the equation into general form.

f(x) = 3(x-2)^2 + 2f(x) = 3(x^2 -4x +4) + 2f(x) = 3x^2 -12x + 12 + 2f(x) = 3x^2 -12x + 14

Now, we can find the derivative of this equation.

f1(x) = 3(2)(x) – (12x^0) - 0= 6x – 12

The exponent on each reduces by one and you multiply the base by the original exponent. The derivative of any constant is zero like mentioned earlier.

Part A ContinuedPart A Continueda) Given the function: f(x) = 3(x-2)^2 + 2, determine the derivative of the function and find the equation of the line tangent to f(x) when x is 4.

So this is the equation to obtain the slope of a line tangent to any point on f(x).

f1(x) = 3(2)(x) – (12x^0) - 0= 6x – 12

We want the line tangent when x=4, so we plug in 4 for x.

f1(4) = 6(4) – 12= 24 – 12= 12

This is just the slope, but we need to find the equation. If you think about it, we can use the point-slope formula

Part A ContinuedPart A Continueda) Given the function: f(x) = 3(x-2)^2 + 2, determine the derivative of the function and find the equation of the line tangent to f(x) when x is 4.

This is the point-slope formula.Y-Y1 = M(X-X1)

The point when x=4 is found on both graphs obviously. However, we need the y-coordinate of the point. What we do is

plug 4 into the original equation, not the derivative one because that equation determines slopes.

f(x) = 3x^2 -12x + 14f(4) = 3(4^2) -12(4) + 14f(4) = 48 – 48 +14f(4) = 14

Therefore, the point is (4,14). Now let’s plug that in.

Part A ContinuedPart A Continueda) Given the function: f(x) = 3(x-2)^2 + 2, determine the derivative of the function and find the equation of the line tangent to f(x) when x is 4.

This is the point-slope formula.

Y – 14 = 12(X-4)Y – 14 = 12x – 48Y = 12x – 48 + 14Y = 12x – 34

This is the equation of the line tangent to f(x) when x=4. Let’s take a look at a graph to see this.

Part A EndPart A Enda) Given the function: f(x) = 3(x-2)^2 + 2, determine the derivative of the function and find the equation of the line tangent to f(x) when x is 4.

You can see that at (4,14), the line Y = 12x – 34 touches f(x) = 3x^2 -12x + 14

Part BPart B

b) The points of intersection between the lines tangent to g(x) when x=4 and the line tangent to f(x) when x=5 are the centers of three circles with a radius of 5. Determine the equations of the circles where g(x)= squareroot[x].

We now have two functions:g(x) = squareroot[x]f(x) = 3(x-2)^2 + 2

Let’s start by finding the line tangent to g(x) when x=4.

Part BPart Bb) The points of intersection between the lines tangent to g(x) when x=4 and the line tangent to f(x) when x=5 are the centers of three circles with a radius of 5. Determine the equations of the circles where g(x)= squareroot[x].

g(x) = squareroot[x]

We have x already, which is 4, so let’s find y.

g(4) = squareroot[4]g(4) = +2 and -2

Oh snap, that’s two values. This is because for every x value on the graph, there are 2 y values, excluding the vertex.

Now let’s obtain the slope

Part BPart Bb) The points of intersection between the lines tangent to g(x) when x=4 and the line tangent to f(x) when x=5 are the centers of three circles with a radius of 5. Determine the equations of the circles where g(x)= squareroot[x].

We must find the derivative of g(x)g(x) = squareroot[x]

g1(x) = (1/2)x^(-1/2)

The squareroot of x is equal to x^(1/2). We multiply that exponent by x and subtract 1 from ½ which gives us -1/2. Now we plug in x to obtain the

slope.

g1(x) = (1/2)x^(-1/2)g1(4) = (1/2)4^(-1/2) *exponents first* *bedmas*g1(4) = (1/2)(1/-2) or (1/2)(1/2)g1(4) = (1/-4) or (1/4)

Okay, so these are our slopes. Now we find the equations.

Part BPart B

What a great slope we have !Slope = +1/4 and -1/4

Now let’s use the point-slope formula to solve for the equation. Earlier, we obtained 2 y values, +2 and -2. We have four possibilities for the equation

now.Y-Y1 = M(X-X1)

Let’s use 2 as our y value with the 2 slopesY – 2 = ¼(x-4)Y = (x/4) + 1

Or

Y – 2 = -¼(x-4)Y = (-x/4) + 3

Let’s use -2 as our y value with the 2 slopesY + 2 = ¼(x-4)Y = (x/4) - 3

Or

Y + 2 = -¼(x-4)Y = (-x/4) - 1

Amazingly, the equations in the black are the correct ones. The next slide shows a graph.

Part BPart B

Part BPart B

We know the derivative of f(x) from earlier, except not the y-value when x=16.

f1(x) = 6x – 12f(x) = 3(x-2)^2 + 2

Let’s find the y-valuef(16) = 3(16-2)^2 + 2f(16) = 3(14)^2 + 2f(16) = 3(16-2)^2 + 2f(16) = 590

Therefore, the point is (16, 590). Now we need to obtain the slope.

b) The points of intersection between the lines tangent to g(x) when x=4 and the line tangent to f(x) when x=5 are the centers of three circles with a radius of 5. Determine the equations of the circles where g(x)= squareroot[x].

Part BPart B

We use the derivative of f(x) to find the slope.f1(x) = 6x – 12

Plug in 16 for Xf1(16) = 6(16) – 12f1(16) = 96 – 12f1(16) = 84Now that we have the slope and 2 points, we can find the equation of

the line.Y-Y1 = M(X-X1)

Y-590 = 84(x-16)Y = 84x – 1344 + 590Y = 84x – 754Great. This is the equation of the line tangent to f(x) when x=16. Let’s

see a graph to verify this.

b) The points of intersection between the lines tangent to g(x) when x=4 and the line tangent to f(x) when x=5 are the centers of three circles with a radius of 5. Determine the equations of the circles where g(x)= squareroot[x].

Part BPart B

The graphs are very ugly, but you can see that the equations are correct.

Now we must find the points of intersection between the tangent lines!

Part BPart BHere are the three equations of tangent lines we have.To find the points of intersection, we make two equations equal other because at that specific point, they share the same coordinate.

For example, making the y’s equal other shows which values of y are shared amongst the graphs.

From g(x), we have:1) Y = (x/4) + 12) Y = (-x/4) - 1

From f(x), we have:3) Y = 84x – 754

There are going to be three intersections because each of these graphs are lines that never end.

I assigned each of the equations numbers to show which intersections I’ll be finding.

Part BPart B1,2

1) Y = (x/4) + 12) Y = (-x/4) – 1

(x/4) + 1 = (-x/4) – 1(x/4) – (-x/4) = – 1 -12x/4 = -2

X = -4Now we must find the y-value to find the coordinate.

We can plug x into anyone of the equations above because they both contain a point when x=-4

Y = (-4/4) + 1Y = 0

Our first point of intersection is (-4,0)

Part BPart B1,3

1) Y = (x/4) + 13) Y = 84x - 754

(x/4) + 1 = 84x – 754(x/4) – (84x) = – 754 – 1(x – 336x)/4 = -755x(1 – 336) = -3020x(-335) = -3020x = 604/67

This is approximately 9.0149 Now we must find the y-value to find the coordinate. We

can plug x into anyone of the equations above because they both contain a point when x=604/67

Y = ((604/67)/4) + 1Y = 151/67 + 67/67Y = 218/67

This approximately equals 3.2537Our second point of intersection is ( 604/67 , 218/67)

Part BPart B

Part BPart B2,3

2) Y = (-x/4) - 13) Y = 84x - 754

(-x/4) - 1 = 84x – 754(-x/4) – (84x) = – 754 + 1(-x – 336x)/4 = -753-337X = -3012x = 3012/337

This approximately equals 8.9377 Now we must find the y-value to find the coordinate. We

can plug x into anyone of the equations above because they both contain a point when x=3012/337

Y = (-(3012/337)/4) + 1Y = -753/337 - 337/337Y = - 1090/337

This approximately equals -3.2344Our third point of intersection is ( 3012/337 , 1090/337 )

Part BPart B

Part BPart BAlright, so our final step is to find the equations of the circles where

their origins are the intersections we found. Each circle has a radius of 5. This is perhaps the HARDEST part of the problem because it

requires so much work to get this far.

Our points of intersection are:

( -4 , 0 ) ( 604/67 , 218/67 )

( 3012/337 , 1090/337 )

The radius is 5.

Now we just plug them into the equation:

(x-h)^2 + (y-k)^2 = r^2

(x+4)^2 + (y-0)^2 = 25(x-604/67)^2 + (y-218/67)^2 = 25

(x-3012/337)^2 + (y-1090/337)^2 = 25

Congratulations!Congratulations!You finished question 3!You finished question 3!