Question

$f\left(x\right)=x^2-7x+10$ find the extreme value

• A. $\frac{-9}{4}$
• B. $\frac{9}{4}$
• C. $\frac{-4}{9}$
• D. $\frac{4}{9}$

Answer 1

The correct option is A $\frac{-9}{4}$

$f\left(x\right)=x^2-7x+10$

The extreme value is at $f^{\prime }\left(x\right)=0$

$f^{\prime }\left(x\right)=2x-7=0x=\frac{7}{2}$

$f\left(\frac{7}{2}\right)={\left(\frac{7}{2}\right)}^2-7\left(\frac{7}{2}\right)+10\frac{49}{4}-\frac{49}{2}+10-\frac{49}{4}+10=\frac{-9}{4}$

The min value of $f\left(x\right)=\frac{-9}{4}$