Browse Questions

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

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

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.