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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find points at which the tangent to the curve \(y = x^3 - 3x^2 - 9x + 7\) is parallel to the \(x\) - axis.

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Toolbox:
  • If $y=f(x)$,then $\big(\large\frac{dy}{dx}\big)_P$=slope of the tangent to $y=f(x)$ at point $P$.
  • If the tangent is parallel to $x$-axis,then $\large\frac{dy}{dx}$$=0$
Step 1:
Given : $y=x^3-3x^2-9x+7$
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}$$=3x^2-6x-9$
Since the tangent to the curve is parallel to the $x$-axis.We get $\large\frac{dy}{dx}$$=0$
$\Rightarrow 3x^2-6x-9=0$
$\Rightarrow 3(x^2-2x-3)=0$
$\Rightarrow 3(x-2)(x+1)=0$
$x=2,-1$
Step 2:
When $x=2$
$y=2^3-3(2)^2-9(2)+7$
$\;\;=8-12-18+7$
$\;\;=-20$
Step 3:
When $x=-1$
$y=(-1)^3-3(-1)^2-9(-1)+7$
$\;\;=12$
Hence the points are $(2,-20)$ and $(-1,12)$
answered Jul 10, 2013 by sreemathi.v
 

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