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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the approximate value of \(f (5.001)\), where $f (x) = x^3 – 7x^2 + 15.$

$\begin{array}{1 1} -34.995 \\ -3.45 \\ -44.995 \\ -36.995 \end{array} $

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1 Answer

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Toolbox:
  • Let $y=f(x)$
  • $\Delta x$ denote a small increment in $x$
  • $\Delta y=f(x+\Delta x)-f(x)$
  • $dy=\big(\large\frac{dy}{dx}\big)\Delta x$
Step 1:
Let $x+\Delta x=5.001$
$\Rightarrow x=5$
$\Delta x=0.001$
$f(x)=f(5)\Rightarrow (5)^3-7(5)^2+15$
$\quad\;\;\;=125-175+15$
$\quad\;\;\;=-35$
$f(x+\Delta x)=f(x)+\Delta f(x)$
$\qquad\;\;\;\;\;\;=f(x)+f'(x).\Delta x$------(1)
Step 2:
Given $f(x)=x^3-7x^2+15$
$f'(x)=3x^2-14x$[Differentiating with respect to x]
Substitute the value of $f(x)$ & $f'(x)$ in equation (1)
$f(x+\Delta x)=(x^3-7x^2+15)+(3x^2-14x).\Delta x$
$f(5.001)=(-35)+(3(5)^2-14\times 5)0.001$
$\quad\;\;\;\;\;\;\;\;\;=(-35)+(75-70)0.001$
$\quad\;\;\;\;\;\;\;\;\;=(-35)+5\times 0.001$
$\Rightarrow -35+5\times (0.001)$
$\Rightarrow -35+0.005$
$\Rightarrow -34.995$
answered Aug 6, 2013 by sreemathi.v
 

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