Question

How do you find the critical numbers for $g\left( t \right) = \left| {3t - 4} \right|$ to determine the maximum and minimum?

Answer 1

In the above question, we are given a function of $t$ as $g\left( t \right) = \left| {3t - 4} \right|$ . We have to find the critical numbers to determine the maximum and minimum of the given function. Critical points of a function are those points for which the value of the derivative of the function is either zero or undefined. First we have to find the derivative of $g\left( t \right)$ that is $g'\left( t \right)$ and then put $g\left( t \right) = 0$ to find the critical point ${t_0}$ .

Complete step-by-step answer:
Given function is
$\Rightarrow g\left( t \right) = \left| {3t - 4} \right|$
We have to find its critical points to determine the maximum and minimum.
Since, $g\left( t \right) = \left| {3t - 4} \right|$ , hence we can also write it in the form
$\Rightarrow g\left( t \right) = \left| {3t - 4} \right|$
Or,

3t - 4,{\text{if }}t \geqslant \dfrac{4}{3} \\\ 4 - 3t,{\text{if }}t < \dfrac{4}{3} \\\ \end{gathered} \right.$$ Now, differentiating the function $$g\left( t \right)$$ once with respect to $$t$$ , we get $$ \Rightarrow g'\left( t \right) = \left\\{ \begin{gathered} 3,{\text{if }}t > \dfrac{4}{3} \\\ \- 3,{\text{if }}t < \dfrac{4}{3} \\\ \end{gathered} \right.$$ Now, note that the derivative $$g'\left( t \right)$$ is either $$3$$ or $$ - 3$$ for the values $$t > \dfrac{4}{3}$$ and $$t < \dfrac{4}{3}$$ but it is not defined for the value $$t = \dfrac{4}{3}$$ . Hence, $$g'\left( t \right)$$ is never equal to zero and also $$g'\left( t \right)$$ is not differentiable at $$t = \dfrac{4}{3}$$ . Hence, here $$t = \dfrac{4}{3}$$ is the only critical point for the function $$g\left( t \right) = \left| {3t - 4} \right|$$ . That means the function $$g\left( t \right)$$ has either minima or maxima on the point $$t = \dfrac{4}{3}$$ . Also the function is a modulus function, hence it is always non-negative. Therefore, it will be minimum only when $$g\left( t \right) = 0$$ at $$t = \dfrac{4}{3}$$ . We can observe that the function $$g\left( t \right)$$ is decreasing on the left of $$t = \dfrac{4}{3}$$ and increasing on the right of $$t = \dfrac{4}{3}$$ . Therefore, $$g\left( t \right)$$ has local minima at $$t = \dfrac{4}{3}$$ . The minimum value is given by $$g\left( {\dfrac{4}{3}} \right)$$ that is $$0$$ . Hence, $$g\left( {\dfrac{4}{3}} \right) = 0$$ is the local minima of $$g\left( t \right) = \left| {3t - 4} \right|$$ at the critical point $$t = \dfrac{4}{3}$$ . **Note:** We can also draw the graph for the function $$g\left( t \right) = \left| {3t - 4} \right|$$ to see the increasing and decreasing nature of the function. Here, we can note that the function has the minimum value at $$t = \dfrac{4}{3}$$ .  The first derivative of a point is the slope of the tangent line at that point. When the slope of the tangent line is 0, the point is either a local minimum or a local maximum. Thus when the first derivative of a point is 0, the point is the location of a local minimum or maximum. The function either has local minima or local maxima at the critical point $${x_0}$$ . If then it has local maxima at $${x_0}$$ and if then the function has local minima at the critical point $${x_0}$$ .