CUET Preparation Today
What is the minimum value of y = \( { x }^{ 2 } -4x\)? |
-4 -8 -2 0 |
-4 |
The correct answer is Option (1) → -4 $y=x^2-4x$ for critical points, $f'(c)=0$ $⇒2c-4=0$ $⇒c=\frac{4}{2}=2$ and, $f''(c)=2>0$ ∴ At $x=2$, $f(x)=y=x^2-4x$ attains it's minimum value. |
