CUET Preparation Today
If x is a real, then minimum value of $x^2-8x +17 $ is : |
-1 0 1 2 |
1 |
The correct answer is Option (3) → 1 $f(x)=x^2-8x +17$ $f'(x)=2x-8$ and for critical point $f'(c)=0$ $⇒2c-8=0$ $⇒c=4$ $f(x)$ is minimum at $x=4$ $f(4)=4×4-8×4+17$ $=33-32=1$ |
