$f(x)=x^2-2x$ has a critical point at
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $x=1$
$f(x)=x^2-2x$
for critical points, $f'(x)=0$
$⇒f'(x)=2x-2=0$
$⇒2x=2$
$⇒x=1$
$f(x)=x^2-2x$ has a critical point at
Correct answer: option 2
The correct answer is Option (2) → $x=1$
$f(x)=x^2-2x$
for critical points, $f'(x)=0$
$⇒f'(x)=2x-2=0$
$⇒2x=2$
$⇒x=1$