Answer to Question #163844 in Algebra for Caren

Question

18. The function f(x) = 3 (x -1/2)² + 4 has

A. Maximum value 4        B. minimum value 4         C. maximum value 4/3       D. minimum value 4/3

Answer 1

$f(x)=3(x-\dfrac{1}{2})^2+4\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space(1)$

Differentiating above equation with respect to x we get

$f^{'}(x)=6(x-\dfrac{1}{2})\times 1+0\space\space\space\space\space\space\space\space\space\space\space\space\space(2)$

Put $f^{'}(x)=0$ in above equation

$\Rightarrow 6(x-\dfrac{1}{2})=0$

$\Rightarrow x=\dfrac{1}{2}$

Put $x=\dfrac{1}{2}$ in equation (1), we get

$\Rightarrow f(\dfrac{1}{2})=4$

Differentiating eq.(2) with respect to x,

$f^{''}(x)=6>0\space(local\space minima\space condition)$

$\therefore \space x=\dfrac{1}{2}\space gives\space the\space minimum\space value\space of\space the\space function\space f(x)$

f(x) = 4 is the minimum value of the function.