# How many horizontal tangents does the graph of f(x)=(x-1)^2(x-3)^2 have?This what I did so far, f(x)=(x-1)^2(x-3)^2 =2(x-1)(1)+2(x-3)(1) =2x-2+2x-6 f'(x)=4x-8 0=4x-8 =4(x-2) x=2 What else am I...

How many horizontal tangents does the graph of f(x)=(x-1)^2(x-3)^2 have?

This what I did so far,

f(x)=(x-1)^2(x-3)^2

=2(x-1)(1)+2(x-3)(1)

=2x-2+2x-6

f'(x)=4x-8

0=4x-8

=4(x-2)

x=2

What else am I suppose to do?

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### 1 Answer

You need to differentiate the function with respect to x, using correctly the product rule such that:

`f'(x)=((x-1)^2)'*(x-3)^2 + (x-1)^2*((x-3)^2)'`

`f'(x) = 2(x-1)(x-3)^2 + 2(x-3)*(x-1)^2`

Notice the difference between the result you will get using correctly the product rule and the result obtained.

You need to solve the equation f'(x) = 0 to find the critical points of function such that:

`2(x-1)(x-3)^2 + 2(x-3)*(x-1)^2 = 0`

You need to factor out `2(x-1)(x-3)` such that:

`2(x-1)(x-3)(x-3+x-1)=0 =gt 2(x-1)(x-3)(2x-4) = 0` `4(x-1)(x-2)(x-3) = 0`

`x - 1 = 0 =gt x = 1`

`x - 2 = 0 =gt x = 2`

`x - 3 = 0 =gt x = 3`

**Hence, the tangent lines to the curve f(x) are parallel to x axis if the slopes are zeroes, hence the number of roots of derivatives gives the number of horizontal tangent lines. Hence,the tangent lines are horizontal at x=1, x=2, x=3.**