Question

Function $f\left(x\right)=x^2-3x+4$ has minimum value of $x$ is

• A. $0$
• B. $-\frac{3}{2}$
• C. $1$
• D. $\frac{3}{2}$

Answer 1

The correct option is D $\frac{3}{2}$

Given, $f\left(x\right)=x^2-3x+4$
$\Rightarrow f^{\prime }\left(x\right)=2x-3$
$\Rightarrow f"\left(x\right)=2$
For critical points. $f^{\prime }\left(x\right)=0$
$\Rightarrow 2x-3=0$
$\Rightarrow x=\frac{3}{2}$
At $x=\frac{3}{2},f"\left(\frac{3}{2}\right)=2>0$
Hence, $f\left(x\right)$ has a minimum value of $x=\frac{3}{2}$.
Therefore, the correct answer from the given alternatives is option $D$.