Question

How do you locate the critical point, and determine the maxima and minima of $y={{x}^{3}}-6{{x}^{2}}+9x-8$? \[\]

Answer 1

Hint: We denote $y=f\left( x \right)={{x}^{3}}-6{{x}^{2}}+9x-8$. We find the critical points $x=c$ as the solutions ${{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=0$. We use the first derivative test to find the maxima and minima. If ${{f}^{'}}\left( x \right)$ changes sign from positive to negative while passing through $x=c$we get maxima at $x=c$ and if ${{f}^{'}}\left( x \right)$ changes sign from negative to positive then we get a minima at $x=c$. \[\]

Complete step by step answer:
We know that minima or maxima of a function occur only at critical points. The critical points of a function are the points where first order derivative of the function zero does not exist. Mathematically, $x=c$ is a critical point if $f'(c)=0$ or $f'(c)$ does not exist. . The first order derivative test is given at critical point $x=c$ as \[\]
1. $f\left( x \right)$ has a local maxima when ${{f}^{'}}\left( x \right)$ changes sign from positive to negative as we increase though the point $x=c$.\[\]
2. $f\left( x \right)$ has a local minima when ${{f}^{'}}\left( x \right)$ changes sign from negative to positive as we increase though the point $x=c$.\[\]
3. $f\left( x \right)$ Will not have neither maxima nor minima if ${{f}^{'}}\left( c \right)=0$ \[\]
Let us denote the given function as $f\left( x \right)$. We have
\[y=f\left( x \right)={{x}^{3}}-6{{x}^{2}}+9x-8\]
We know that polynomial functions are differentiable everywhere in $R$. Here the given function is a cubic polynomial function and hence is differentiable everywhere in $R$. So the critical points will be solutions of ${{f}^{'}}\left( x \right)=0$. We differentiate the given function to have;
 \[\begin{align}
  & {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{x}^{3}}-6{{x}^{2}}+9x-8 \right) \\
 & \Rightarrow {{f}^{'}}\left( x \right)=3{{x}^{2}}-12x+9 \\
 & \Rightarrow {{f}^{'}}\left( x \right)=3\left( {{x}^{2}}-4x+3 \right) \\
\end{align}\]
We equate the first derivative to zero to have;
\[\begin{align}
  & {{f}^{'}}\left( x \right)=0 \\
 & \Rightarrow {{x}^{2}}-4x+3=0 \\
 & \Rightarrow {{x}^{2}}-3x-x+3=0 \\
 & \Rightarrow x\left( x-3 \right)-1\left( x-3 \right)=0 \\
 & \Rightarrow \left( x-3 \right)\left( x-1 \right)=0 \\
 & \Rightarrow x=3,x=1 \\
\end{align}\]
So the critical points are $x=1,3$. Let us take three points so $x=0,x=2,x=4$ in the left of $x=1$ , in the interval $\left( 1,3 \right)$ and in the right of $x=3$ respectively. Let us find the first derivative at $x=0,x=2,x=4$.
\[\begin{align}
  & {{f}^{'}}\left( 0 \right)=3\left( {{0}^{2}}-4\cdot 0+3 \right)=9>0 \\
 & {{f}^{'}}\left( 2 \right)=3\left( {{2}^{2}}-4\cdot 2+3 \right)=-3<0 \\
 & {{f}^{'}}\left( 4 \right)=3\left( {{4}^{2}}-4\cdot 4+3 \right)=9>0 \\
\end{align}\]
We see that the function changes its sign from positive to negative when it passes through $x=1$ and changes its sign from negative to positive when it passes through $x=3.$So the there is a maxima at $x=1$ and the minima at $x=3$.So we have;
\[\begin{align}
  & \text{maxima: }f\left( 1 \right)={{1}^{3}}-6\cdot {{1}^{2}}+9\cdot 1-8=-4 \\
 & \text{maxima: }f\left( 3 \right)={{3}^{3}}-6\cdot {{3}^{2}}+9\cdot 3+8=-8 \\
\end{align}\]

Note:
We note that if we get ${{f}^{'}}\left( x \right)$ does not change sign at critical points the first derivative test fails there and we move on to second derivative test which states that there is a minima at $x=c$ if ${{f}^{''}}\left( c \right)>0$ ad there is a maxima at $x=c$ if ${{f}^{''}}\left( c \right)>0$. We also note that derivative tests give us local maxima or minima not global maxima or minima.