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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives

If x is real,the minimum value of $x^2-8x+17$ is

\[(A)\;-1\quad(B)\;0\quad(C)\;1\quad(D)\;2\]

1 Answer

Toolbox:
  • To obtain the absolute maxima or minima for the function $f(x)$
  • (i) Find $f'(x)$ and put $f'(x)=0$
  • (ii) Obtain the points from $f'(x)=0$
  • (iii) By phthogoras theorem, find $f''(x)$ and check the value of $f''(x)$ for each of the points obtained if $f''(x)>0$ it has a minimum at that point. If $f''(x)<0$ then it has a maximum at that point.
Step 1
$y=x^2-8x+17$
On differentiating w.r.t $x$ we get,
$ \large\frac{dy}{dx}=2x-8$
Again differentiating w.r.t $x$ we get,
$ \large\frac{d^2y}{dx^2}=2$
Since $ \large\frac{d^2y}{dx^2}>0,$ the value should be a minimum.
Step 2
To obtain the minimum value,
when $ \large\frac{dy}{dX}=0$
$ \Rightarrow 2x-8=0$
$ \Rightarrow x=4$
when $ x=4, \: y=(4)^2-8(4)+17$
$=1$
The minimum value is 1
Hence C is the correct option.
answered Aug 11, 2013 by thanvigandhi_1
 
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