Question

Suppose that the cubic function f(x)=(x-a)(x-b)(x-c) has three distinct zeroes: a, b, c. Prove that a tangent line drawn at the average of the zeroes a and b intersect the graph of f at the third zero?

Answer 1

See below.

Explanation:

Given

`f(x) = (x-a)(x-b)(x-c)`

`x_0 = (a+b)/2`

the tangent line at `x_0` is

`y = y_0 + (deltaf_0)(x-x_0)`

where

`y_0 = f(x_0) = 1/8(a-b)^2(2c-a-b)`
`delta f_0 = f'(x_0) = 3x^2-2(a+b+c)x_0+ab+ac+bc = -1/4(a-b)^2`

now the intersection between the tangent line and `f(x)` we have to solve

`f(x) = y_0 + (deltaf_0)(x-x_0)` or

`x^3-(a+b+c)x^2+(ab+ac+bc)x-abc -( 1/8(a-b)^2(2c-a-b)-1/4(a-b)^2(x-(a+b)/2))=0`

or after simplifications

`1/2(a+b-2x)^2(x-c)=0`

and as we can observe `x = c` is a root as expected.

Answer 2

See below:

Explanation:

`f((a+b)/2)=((a+b)/2-a)((a+b)/2-b)((a+b)/2-c)`
`qquad =-1/4(a-b)^2((a+b)/2-c)`

Thus, the slope of the line joining the point `((a+b)/2,f((a+b)/2))` to the point `(c,0)` is

`(f((a+b)/2)-0)/((a+b)/2-c) = -1/4(a-b)^2`

Now, the derivative of `f(x)` is given by

`f'(x)=(x-a)(x-b)+(x-b)(x-c)+(x-a)(x-c) = (x-a)(x-b)+(2x-a-b)(x-c)`

So, the slope of the tangent to the curve at the point `((a+b)/2,f((a+b)/2))` is

`f'((a+b)/2)= ((a+b)/2-a)((a+b)/2-b)+(2 times (a+b)/2-a-b)((a+b)/2-c) = -1/4(a-b)^2`

Thus the tangent at the point `((a+b)/2,f((a+b)/2))` goes through `(c,0)`.