To determine

(a)

To sketch: The graph of f and its tangent lines at the average of each pairs of zeros and to explain what will notice

Answer

The graph of f and its tangent lines at the average of each pairs of zeros

Explanation

1) Concept:

Take the average of each pair of zeros and then draw the tangent line at that point and then conclude.

2) Given:

fx=x(x-2)(x-6)

3) Graph:

Consider the purple tangent line at

x=2+62=4,

X intercept of this line is at x = 0 which is third zero of the cubic function

Now, consider blue tangent at

x=0+62=3

X intercept of this line is at x = 2 which is third zero of the cubic function

Now, consider black tangent at

x=0+22=1

X intercept of this line is at x = 6 which is third zero of the cubic function

Therefore, it will notice that the graph of f and its tangent lines at the average of each pairs of zeros intersects the graph at third zero

Conclusion:

The graph of f and its tangent lines at the average of each pairs of zeros

It will notice that the graph of f and its tangent lines at the average of each pairs of zeros intersects the graph at third zero

To determine

(b)

To prove:

The tangent line drawn at the average of zeros a and b intersects the graph of f at the third zero

Answer

The tangent line drawn at the average of zeros a and b intersects the graph of f at the third zero

Formula:

i) Constant multiple rule:

ddxCfx=Cddx(fx)

ii) Sum rule:

ddxfx+gx=ddxfx+ddx(gx)

iii) Difference rule:

ddxfx-gx=ddxfx-ddx(gx)

iv) Power rule:

ddxxn=nxn-1

v) Constant function rule:

ddxC=0

vi) a+b2=a2+2ab+b2

vii) Point-slope formula:

y-y1=m(x-x1)

Explanation

1) Concept:

Draw the tangent at average of zeros a and b and calculate slope at the same point and find x intercept from the equation of tangent line which will be third zero of the given cubic equation.

2) Calculations:

fx=(x-a)(x-b)(x-c)

We have to draw tangent line at a+b2 that is at the average of zeros a and b and to show it is c

First find the y coordinate,

fa+b2=a+b2-aa+b2-ba+b2-c

⇒fa+b2=a+b-2a2a+b-2b2a+b-2c2

⇒fa+b2=b-a2a-b2a+b2-c

⇒fa+b2=-14a-b2a+b2-c

Now consider the function,

fx=(x-a)(x-b)(x-c)

By simplifying it,

fx=(x2-xb-ax+ab)(x-c)

⇒fx=x3-bx2-ax2+abx-cx2+cbx+cax-abc

⇒fx=x3-(a+b+c)x2+(ab+cb+ca)x-abc

Now differentiate with respect to x,

f'x=ddx[x3-(a+b+c)x2+(ab+cb+ca)x-abc]

By using sum and difference rule,

f'x=ddxx3-ddx(a+b+c)x2+ddx(ab+cb+ca)x-ddx(abc)

By using constant multiple rule and constant function rule,

f'x=ddxx3-(a+b+c)ddxx2+(ab+cb+ca)ddxx-0

By using power rule,

f'x=3x2-(a+b+c)(2x)+(ab+cb+ca)

At  x=a+b2

f'a+b2=3a+b22-a+b+c2a+b2+(ab+cb+ca)

f'a+b2=34(a2+2ab+b2)-(a+b+c)(a+b)+(ab+cb+ca)

f'a+b2=34(a2+2ab+b2)-(a2+ab+ac+ab+b2+bc)+(ab+cb+ca)

f'a+b2=34a2+2ab+b2-a2-ab-ac-ab-b2-cb+ab+cb+ca

f'a+b2=34a2+2ab+b2-a2-ab-b2

f'a+b2=3a2+6ab+3b2-4a2-4ab-4b24

f'a+b2=-a2+2ab-b24

⇒f'a+b2=-14a-b2

By using point-slope formula,

y+14a-b2a+b2-c=-14a-b2(x-a+b2)

At x- intercept  y = 0

⇒14a-b2a+b2-c=-14a-b2(x-a+b2)

⇒a+b2-c=-(x-a+b2)

⇒a+b2-c=-x+a+b2

⇒-c=-x

⇒x=c

Hence the proof

Conclusion:

The tangent line drawn at the average of zeros a and b intersects the graph of f at the third zero