Question

What is the min and the max value of y=(2x-1)^3+3 by the derivative method

Answer 1

Given,

$\longrightarrow\sf{y=(2x-1)^3+3\quad\quad\dots(1)}$

For finding value of $\sf{x}$ at which $\sf{y}$ is maximum or minimum, its derivative should be equated to zero.

$\longrightarrow\sf{y'=0}$

$\longrightarrow\sf{\dfrac{d}{dx}\,[(2x-1)^3+3]=0}$

$\longrightarrow\sf{\dfrac{d[(2x-1)^3]}{d[2x-1]}\cdot\dfrac{d[2x-1]}{dx}+\dfrac{d}{dx}(3)=0}$

$\longrightarrow\sf{3(2x-1)^2\cdot2=0}$

$\longrightarrow\sf{6(2x-1)^2=0}$

$\longrightarrow\sf{(2x-1)^2=0}$

$\longrightarrow\sf{2x-1=0}$

$\longrightarrow\sf{x=\dfrac{1}{2}}$

Since $\sf{x}$ has only one value, $\sf{y}$ has only one stationary point hence it has no maxima nor minima.

From (1),

$\longrightarrow\sf{y=\left(2\times\dfrac{1}{2}-1\right)^3+3}$

$\longrightarrow\sf{y=3}$

Hence the stationary point is $\bf{\left(\dfrac{1}{2},\ 3\right)}$