Recently I wrote about using the First Derivative (FD) Test to find the intervals where a function is increasing or decreasing. With the FD Test, we can also find the exact value where the function is highest or lowest in an interval (called absolute maxima) or, where it is higher or lower than neighboring values (called relative maxima). I will be using 2 different functions.
Function #1
\[f\left(x\right) = \left(x-1\right)^{2}\left(3+x\right), \quad -2 \le x \le 2\]
It would be simple to expand the function before differentiating, but what to do if instead of \(\left(x-1\right)^{2}\), we had \(\left(x-1\right)^{100}\)? Expanding this binomial would take a long time and be prone to errors. So, let’s take the derivative of the original function using the Product and Chain Rules and simplify the result:
\begin{alignat*}{1}
\frac{d}{dx}f\!\left(x\right) &= \left(x-1\right)^{2}+\left(x+3\right)[\left(2\right)\left(x+1\right)]\\
&= 3x^{2}+2x-5
\end{alignat*}
At this point, we set the derivative function to zero and solve for \(\boldsymbol x\).
\begin{alignat*}{1}
3x^{2}+2x-5 &= 0\\
\left(3x+5\right)\left(x-1\right) &= 0\\
x &= -\frac{5}{3},1 \quad \leftarrow \text{Critical Numbers}
\end{alignat*}
These values of \(\boldsymbol x\) are called Critical Numbers; they are where the slope of the tangent line to the function is zero and could be where the function ‘turns’. To verify this, we select a suitable value within each interval to check the sign of the slope: \(\left(-2,-\frac{5}{3}\right), \left(-\frac{5}{3},1\right)\), and \(\left(1,2\right)\).
\[f’\!\left(-1.99\right) \gt 0, \qquad f’\!\left(0\right) \lt 0, \qquad f’\!\left(1.5\right) \gt 0\]
Because the slope is positive, we see the function is increasing in the interval \(\left(-2,-\frac{5}{3}\right)\), decreasing \(\left(-\frac{5}{3},1\right)\) because the slope is negative in that interval, and increasing again in \(\left(1,2\right)\) since the slope is positive then.
Where is the function \(\boldsymbol{f\!\left(x\right)}\) highest and lowest? To find out we try the 2 critical points, and because we are looking at the close interval \([-2,2]\), also the endpoints.
\begin{alignat*}{1}
f\!\left(-\frac{5}{3}\right) &= 9.4815\\
f\!\left(1\right) &= 0\\
f\!\left(-2\right) &= 9\\
f\!\left(2\right) &= 5
\end{alignat*}
From the results above, we see that the absolute maximum is at \(\left(-\frac{5}{3},9.4815\right)\), absolute minimum is at \(\left(1,0\right)\), as the following graph confirms.
If you find any errors in this post, or have some thoughts about it, please, email me at [email protected] .