Monotonicity, Maxima and Minima
7.0 First Derivative Test
7.0 First Derivative Test
It is the standard method to find the points of local maxima or local minima of the given function. Let us consider a continuous function $f(x)$ with a critical point at $x=c$ in an open interval $I$.
To find whether $'c'$ is the point of local maxima or local minima, just differentiate the function with respect to $x$, we get $f'(x)$. Then, we check the sign of $f'(x)$ just left to the point $'c'$ and just right to it.
- $'c'$ be a point of local maxima if $f'(x)$ changes sign from positive to negative as the value of $x$ increases through $'c'$. In simple words, we can say that if the sign of $f'(x)$ at the values less than $'c'$ is positive and at the values greater than $'c'$ is negative, then $x=c$ is the point of local maxima.
- $'c'$ be a point of local minima if $f'(x)$ changes sign from negative to positive as the value of $x$ increases through $'c'$. In simple words, we can say that if the sign of $f'(x)$ at the values less than $'c'$ is negative and at the values greater than $'c'$ is positive, then $x=c$ is the point of local maxima.
- $'c'$ be neither a point of maxima or minima if $f'(x)$ does not change sign as the value of $x$ increases through $'c'$. In simple words, we can say that if the sign of $f'(x)$ at the values less than $'c'$ is same as at the values greater than $'c'$, then $x=c$ is neither a point of maxima or minima. If this is the case, then check for the point of inflection using the steps explained above.
Question 13. Find all the points of local maxima and local minima of the function $$f(x) = 2{x^3} - 6{x^2} + 6x + 5$$
Solution: Differentiating the function, we get $$\begin{equation} \begin{aligned} f(x) = 2{x^3} - 6{x^2} + 6x + 5 \\ f'(x) = 6{x^2} - 12x + 6 \\ f'(x) = 6{x^2} - 6x - 6x + 6 \\ f'(x) = 6x(x - 1) - 6(x - 1) \\ f'(x) = 6{(x - 1)^2} = 0 \\ \Rightarrow x = 1 \\\end{aligned} \end{equation} $$ Therefore, $x=1$ is the critical point. Now, we should check the sign of $f'(x)$ just before and after $x=1$. Using number line, we can say that $$f'(x) > 0\;;\;x \in R$$ Using first derivative test, we can say that $x=1$ is neither a point of maxima nor a point of minima. Now, we check the sign of $f''(x)$ i.e., $$\begin{equation} \begin{aligned} f''(x) = 12(x - 1) \\ \Rightarrow f''(x) > 0,{\text{ }}x > 1{\text{ and }}f''(x) < 0,{\text{ }}x < 1 \\\end{aligned} \end{equation} $$ Since, $f''(x)$ changes sign at $x=1$, we can say that it is a point of inflection.
