How many horizontal tangents does the graph of f(x)=(x-1)^2(x-3)^2 have?
This what I did so far,
What else am I suppose to do?
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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.
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