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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the point on the curve \(y = x^3 - 11x + 5\) at which the tangent is \(y = x -11\).

$\begin{array}{1 1} 2,9 \\ 2,-9 \\ -2,9 \\ -2,-9 \end{array} $

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1 Answer

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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 the point $P$.
  • Equation of line with given two points is $\large\frac{x-x_1}{x_2-x_1}=\large\frac{y-y_1}{y_2-y_1}$
Step 1:
Given : $y=x^3-11x+5$
Differentiate w.r.t $x$ we get,
$\large\frac{dy}{dx}$$=3x^2-11$----(1)
The equation of the given tangent is $y=x-11$
Step 2:
Hence the slope of the tangent is $1$.
Now equating this slope to equ(1)
$3x^2-11=1$
$3x^2=12$
$x^2=4$
$x=\pm 2$
Step 3:
When $x=2,y=-9$
When $x=-2,y=-13$
Hence the point is $(2,-9)$
answered Jul 10, 2013 by sreemathi.v
 

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