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Question
At what point on the following curve, is the tangent parallel to x-axis y = 12 (x + 1) (x − 2) on [−1, 2] ?
Solution
Let \[f\left( x \right) = 12\left( x + 1 \right)\left( x - 2 \right)\] ...(1)
Since \[f\left( x \right)\] is a polynomial function, \[f\left( x \right)\] is continuous on \[\left[ - 1, 2 \right]\] and differentiable on \[\left( - 1, 2 \right)\].
Consequently, there exists at least one point c \[\in \left( - 1, 2 \right)\] for which \[f'\left( c \right) = 0\] .
By the geometrical interpretation of Rolle's theorem, \[\left( \frac{1}{2}, - 27 \right)\] is the point on \[y = 12\left( x + 1 \right)\left( x - 2 \right)\] where the tangent is parallel to the x-axis.
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