PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Solution for At What Point on the Following Curve, is the Tangent Parallel to X-axis Y = 12 (X + 1) (X − 2) on [−1, 2] ? - PUC Karnataka Science Class 12 - Mathematics

ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### 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)

$\Rightarrow$ $f\left( x \right) = 12\left( x^2 - x - 2 \right)$
$\Rightarrow$ $f\left( x \right) = 12 x^2 - 12x - 24$

Since

$f\left( x \right)$ is a polynomial function, $f\left( x \right)$ is continuous on $\left[ - 1, 2 \right]$ and differentiable on $\left( - 1, 2 \right)$.
Also,
$f\left( 2 \right) = f\left( - 1 \right) = 0$
Thus, all the conditions of Rolle's theorem are satisfied.
Consequently, there exists at least one point $\in \left( - 1, 2 \right)$ for which $f'\left( c \right) = 0$ .
But
$f'\left( c \right) = 0 \Rightarrow 24c - 12 = 0 \Rightarrow c = \frac{1}{2}$
$\therefore f\left( c \right) = f\left( \frac{1}{2} \right) = - 12\left( \frac{3}{2} \right)\left( \frac{3}{2} \right) = - 27$ (using (1))
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.
Is there an error in this question or solution?

#### APPEARS IN

Solution At What Point on the Following Curve, is the Tangent Parallel to X-axis Y = 12 (X + 1) (X − 2) on [−1, 2] ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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