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# Derivative test – Types, Explanation, and Examples

**Derivative tests**are applications of derivatives that help us determine whether a critical number is a local maximum or minimum. Learning about the

**first and second derivative tests**will help us confirm a critical number’s nature without graphing the actual function.

**By the end of this article, you should be able to:**

*The first and second derivative tests allow us to determine a function’s potential critical numbers and local extrema. These two tests tell us whether the function is concaving upward or downward within a given interval.*- Finding the critical numbers of a function using its first derivative.
- Use the two derivative tests to confirm if a critical number is a local extremum.
- Know the difference between the two derivative tests.

**What is the first derivative test?**

The first derivative test allows us to find the intervals where the **function is increasing or decreasing.**This derivative also helps us

**find the local maxima and minima of a function**. According to the first derivative test, when $c$ is a critical number of $f(x)$, we can observe the following:

- When $f’(x)$ switches
**from negative to positive**at $c$, $f(x)$ has a**local (or relative) minimum**at $(c, f(c))$. - Similarly, when $f’(x)$ switches from
**positive to negative**at $c$, $f(x)$ has a**local (or relative) maximum**at $(c, f(c))$. - If the sign of $f’(x)$
**remains the same**on both sides of $c$, $f(x)$ has**no extremum**at $(c, f(c))$.

- At $x = -\dfrac{8}{3}$, we can see that $f’(x)$’s sign changes from positive to negative, so we have a local maximum at $x = -\dfrac{8}{3}$.
- Similarly, $f’(x)$’s sign changes from negative to positive at $x =0$, so we have a local minimum at $x = 0$.

**How to do the first derivative test? **

We will still use the conditions discussed in our earlier discussion of the first derivative test. These steps will guide you in applying this method correctly:
- Find the critical numbers of $x$ by equating $f’(x)$ to $0$.
- Divide the domain of $f(x)$ into smaller intervals containing the critical numbers as bounds.
- Use a test point for each interval and observe the sign of $f’(x)$.
- Use these results and the first derivative test’s conditions to determine whether the function has a local maximum or minimum.

Interval |
Test Value |
Sign of $\boldsymbol{f’(x)}$ |
Conclusion |

\begin{aligned}\boldsymbol{(-\infty, 2)}\end{aligned} | \begin{aligned}x&=1\end{aligned} | \begin{aligned}-\end{aligned} | $f(x)$ is decreasing |

\begin{aligned}\boldsymbol{(2, \infty)}\end{aligned} | \begin{aligned}x&=3\end{aligned} | \begin{aligned}+\end{aligned} | $f(x)$ is increasing |

**What is the second derivative test?**

The second derivative test comes in handy when we have twice differentiable functions (meaning, we can take the derivative $f(x)$ twice in a row). We can confirm the relative maxima using $f”(x)$.
- When $\boldsymbol{f”(x) >0}$, the function has a
**local minimum**at $(c, f(c))$. - When $\boldsymbol{f”(x) <0}$, the function has a
**local maximum**at $(c, f(c))$. - When $\boldsymbol{f”(x) =0}$, the second derivative test
**will not be helpful**. Use the first derivative test instead.

- At $x =-1.02$, the double derivative of $f(x)$ is negative, so $(-1.02, 1.08)$ is the function’s relative maximum.
- Similarly, since $f”(x) > 0$ at $x =1.10$, the point $(1.10, -1.05)$ is a relative minimum.

**is less than zero**, the function is

**concaving downwards**. Similarly, when $f”(x)$

**is greater than zero**, the function is

**concaving upward**.

**How to do the second derivative test?**

The good thing about the second derivative test is that we’ll only evaluate $f”(x)$ at $c$ as opposed to the first derivative test. Keep these steps in mind when applying the second derivative test to check for relative extremums:
- Differentiate $f(x)$ and equate $f’(x)$ to zero to find the function’s critical numbers.
- Differentiate $f’(x)$ this time to find the second derivative of $f(x)$.
- Substitute $x = c$ or the critical numbers of $f(x)$ into the expression of $f”(x)$.
- Observe the signs and use the conditions we’ve learned to confirm whether $x=c$ is a local maximum or minimum.

**All values that satisfy this equation will be considered a critical number**. \begin{aligned}f'(x) &=0\\5x^4-6x^2&=0\\x^2(5x^2 -6) &=0\\x&=0\\5x^2 – 6&=0\\5x^2&= 6\\x^2&= \dfrac{6}{5}\\x&= \pm \sqrt{\dfrac{6}{5}}\end{aligned} Now, differentiate $f’(x)$ once more and then substitute these three values into the resulting expression. \begin{aligned}f'(x) &=5x^4-6x^2\\f”(x)&=\dfrac{d}{dx}5x^4 – \dfrac{d}{dx}6x^2,\phantom{x}\color{Teal}\text{Difference Rule}\\&= {\color{Teal}5\dfrac{d}{dx}x^4} -{\color{Teal}6\dfrac{d}{dx}x^2},\phantom{x}{\color{Teal}\text{Constant Multiple Rule}}\\&=5{\color{Teal}(4x^3)}-6{\color{Teal}(2x)},\phantom{x}{\color{Teal}\text{Power Rule}}\\&= 20x^3-12x\end{aligned} The table below summarizes the result when we substitute $x =\left\{-\sqrt{\dfrac{6}{5}},0,\sqrt{\dfrac{6}{5}}\right\}$ into $f”(x)$.

Critical Numbers |
Sign of $\boldsymbol{f’’(x)}$ |
Conclusion |

\begin{aligned}x = -\sqrt{\dfrac{6}{5}}\end{aligned} | \begin{aligned}-\end{aligned} | Local Maximum |

\begin{aligned}x = 0\end{aligned} | \begin{aligned}0\end{aligned} | No conclusion |

\begin{aligned}x =\sqrt{\dfrac{6}{5}}\end{aligned} | \begin{aligned}+\end{aligned} | Local Minimum |

**Summarizing the two derivative tests**

We’ve discussed the first and second derivative tests, so we’ll summarize and do a quick recap of what we’ve learned so far.
- In the first derivative test, $(c, f(c))$ is a local extremum when $f’(x)$ changes sign.
- When $f’(x)$’s sign changes from negative to positive, $(c, f(c))$ is a local minimum.
- When $f’(x)$’s sign changes from positive to negative, $(c, f(c))$ is a local maximum.

- As for the second derivative test, $(c, f(c))$ will depend on $f”(c)$’s sign.
- When $f”(c)<0$, the point $(c, f(c))$ is a local maximum.
- When $f”(c)>0$, the point $(c, f(c))$ is a local minimum.

Observing the First and Second Derivative’s Signs |
Function’s Behavior |
Shape of Function’s Curve |

\begin{aligned}f’(x)&>0\\ f”(x) &> 0\end{aligned} | $f(x)$ is increasing | $f(x)$ is concaving upward |

\begin{aligned}f’(x)&<0\\ f”(x) &> 0\end{aligned} | $f(x)$ is decreasing | $f(x)$ is concaving upward |

\begin{aligned}f’(x)&>0\\ f”(x) &< 0\end{aligned} | $f(x)$ is increasing | $f(x)$ is concaving downward |

\begin{aligned}f’(x)&<0\\ f”(x) &> 0\end{aligned} | $f(x)$ is decreasing | $f(x)$ is concaving downward |

**Let’s say we have the function, $f(x) = \dfrac{x^2 – 6x +9}{x^2 -4}$, find the following: a. The critical numbers of $f(x)$. b. The local extrema of $f(x)$ using the first derivative test. c. The intervals where $f(x)$ is increasing and decreasing.**

*Example 1*__Solution__We begin by differentiating $\dfrac{x^2 – 6x +9}{x^2 -4}$ using the quotient rule and other derivative rules as shown below. \begin{aligned}f(x) &=\dfrac{x^2 -6x +9}{x^2 -4}\\f'(x)&=\dfrac{(x^2 -4)\dfrac{d}{dx}(x^2-6x+9)-(x^2-6x+9)\dfrac{d}{dx}(x^2 -4)}{(x^2 -4)^2},\phantom{x}\color{Teal}\text{Quotient Rule}\\&=\dfrac{(x^2 -4){\color{Teal}\left(\dfrac{d}{dx}x^2-\dfrac{d}{dx}6x+\dfrac{d}{dx}9\right)}-(x^2-6x+9){\color{Teal}\left(\dfrac{d}{dx}x^2 -\dfrac{d}{dx}4\right)}}{(x^2 -4)^2},\phantom{x}\color{Teal}\text{Sum & Difference Rules}\\&=\dfrac{(x^2 -4)\left(\dfrac{d}{dx}x^2-{\color{Teal}6\dfrac{d}{dx}x}+{\color{Orchid}0}\right)-(x^2-6x+9)\left(\dfrac{d}{dx}x^2 -{\color{Orchid}0}\right)}{(x^2 -4)^2},\phantom{x}{\color{Teal} \text{Constant Multiple Rule}}\text{ & }{\color{Orchid}\text{Constant Rule}}\\&= \dfrac{(x^2 -4)[{\color{Teal}2x} – 6({\color{Teal}1})]-(x^2-6x+9)({\color{Teal}2x} )}{(x^2 -4)^2},\phantom{x}\color{Teal}\text{Power Rule}\end{aligned} Simplify the expression for $f’(x)$ by algebraic manipulation then equate the resulting expression to $0$. This will lead us to the value of $x$ where $f(x)$ is an extremum. \begin{aligned}f'(x)&= \dfrac{(x^2-4)(2x -6)-(x^2-6x+9)(2x)}{(x^2 -4)^2}\\&= \dfrac{2(x^2 -4)(x-3) – (x-3)^2(2x)}{(x^2-4)^2}\\&= \dfrac{2(x-3)[(x^2 -4)-x(x-3)]}{(x^2 -4)^2}\\&= \dfrac{2(x-3)(3x-4)}{(x^2-4)^2}\\\\f'(x)&= 0\\\dfrac{2(x-3)(3x -4)}{(x^2 -4)^2} &= 0\\2(x -3)(3x-4)&=0\\x&= 3\\x&= \dfrac{4}{3}\end{aligned} a. Hence, the critical numbers of $f(x)$ are $x = \dfrac{4}{3}$ and $x =3$. As long as $x \neq \pm 2$, $f(x)$ is valid and continuous. Divide the domain of $f(x)$, $(-\infty, -2) \cup(-2, 2)\cup (2, \infty)$, into smaller intervals to account for the critical numbers. Observe the $f’(x)$’s sign changes between these intervals.

Interval |
Test Value |
Sign of $\boldsymbol{f’(x)}$ |
Conclusion |

\begin{aligned}\left (-\infty, -2\right)\end{aligned} | \begin{aligned}x&=-3\end{aligned} | \begin{aligned}+\end{aligned} | $f(x)$ is increasing |

\begin{aligned} \left(-2, -\dfrac{4}{3}\right)\end{aligned} | \begin{aligned}x&=-1.5\end{aligned} | \begin{aligned}+\end{aligned} | $f(x)$ is increasing |

\begin{aligned} \left(-\dfrac{4}{3}, 2\right)\end{aligned} | \begin{aligned}x&=1\end{aligned} | \begin{aligned}-\end{aligned} | $f(x)$ is decreasing |

\begin{aligned} (2, 3)\end{aligned} | \begin{aligned}x&=2.5\end{aligned} | \begin{aligned}-\end{aligned} | $f(x)$ is decreasing |

\begin{aligned} (3, \infty)\end{aligned} | \begin{aligned}x&=4\end{aligned} | \begin{aligned}+\end{aligned} | $f(x)$ is increasing |

**Confirm that the function from Example 1 has a local maximum at $x = \dfrac{4}{3}$ and a local minimum at $x = 3$ using the second derivative test. Use these results to determine the intervals where $f(x)$ is concaving upwards and downwards.**

*Example 2*__Solution__From Example 1, we have $f’(x) = \dfrac{2(x-3)(3x -4)}{(x^2 -4)^2}$. We’ve also shown that $x = \dfrac{4}{3}$ and $x = 3$ are $f(x)$’s critical numbers. We can confirm the local extremums of $f(x)$ by taking the derivative of $f’(x)$. First, let’s simplify the numerator of $f’(x)$ then apply the derivative rules as shown below. \begin{aligned}f'(x)&= \dfrac{2(x-3)(3x -4)}{(x^2 -4)^2} \\&= \dfrac{6x^2 – 26x + 24}{(x^2 -4)^2}\\\\f”(x)&= \dfrac{(x^2 -4)^2 \dfrac{d}{dx}(6x^2 – 26x + 24) – (6x^2 – 26x + 24)\dfrac{d}{dx}(x^2-4)^2}{(x^2 -4)^4},\phantom{x}{\color{Teal}\text{Quotient Rule}}\\&= \dfrac{(x^2 -4)^2 {\color{Teal}\left(\dfrac{d}{dx}6x^2 – \dfrac{d}{dx}26x + \dfrac{d}{dx}24\right)}-(6x^2-26x + 24)\dfrac{d}{dx}(x^2 -4)^2}{(x^2 -4)^4},\phantom{x}{\color{Teal}\text{Sum & Difference Rules}}\\&= \dfrac{(x^2 -4)^2 \left(\dfrac{d}{dx}6x^2-\dfrac{d}{dx}26x + \dfrac{d}{dx}24\right) – (6x^2 – 26x + 24){\color{Teal}2(x^2-4)}{\color{Orchid}\dfrac{d}{dx}(x^2 -4)}}{(x^2 -4)^4},\phantom{x}{\color{Teal}\text{Power Rule}}\text{ & }{\color{Orchid}\text{Chain Rule}}\\&= \dfrac{(x^2-4)^2\left({\color{Teal}6\dfrac{d}{dx}x^2}-{\color{Teal}26\dfrac{d}{dx}x }+\dfrac{d}{dx}24 \right )-2(x^2-4)(6x^2-26x+24)\left(\dfrac{d}{dx}x^2 -\dfrac{d}{dx}4\right)}{(x^2-4)^4},\phantom{x}{\color{Teal}\text{Constant Multiple Rule}}\end{aligned} Simplify the expression in the numerator further to find $f”(x)$. \begin{aligned}f”(x)&= \dfrac{(x^2 -4)^2[6{\color{Teal}(2x)}-26{\color{Teal}(1) }+{\color{Orchid}0} ]-2(x^2-4)(6x^2-26x+24)[{\color{Teal}(2x)}-{\color{Orchid}0} ]}{(x^2 -4)^4},\phantom{x}{\color{Teal}\text{Power Rule}}\text{ & }{\color{Orchid}\text{Constant Rule}}\\&= \dfrac{(12x-26)(x^2 -4)^2-4x(x^2-4)(6x^2-26x+24) }{(x^2-4)^4}\\&= \dfrac{2(x^2 -4)[(6x-13)(x^2 -4)-2x(6x^2-26x+24)] }{(x^2 -4)^4}\\&= \dfrac{2[(6x-13)(x^2 -4)-2(6x^2-26x+24)] }{(x^2 -4)^3}\\&= -\dfrac{2(6 x^3-39 x^2+72 x-52)}{(x^2 -4)^3}\end{aligned} You may use a calculator or graphing utility to estimate the values of $f”(x)$ at $x = \dfrac{4}{3}$ and $x = 3$.

Critical Numbers |
Sign of $\boldsymbol{f’’(x)}$ |
Conclusion |

\begin{aligned}x =\dfrac{4}{3}\end{aligned} | \begin{aligned}-\end{aligned} | Local Maximum, $f(x)$ is concaving downward |

\begin{aligned}x =3 \end{aligned} | \begin{aligned}+\end{aligned} | Local Minimum, $f(x)$ is concaving upward |

**Practice Questions**

1. Find the local extrema of the following functions using the first derivative test:
a.$f(x)=4x^3 -6x$
b.$g(x)=-4x^4-1$
c.$h(x)= \dfrac{x}{x-6}$
2. Find the local extrema of the following functions using the second derivative test:
a.$f(x)=-6x^5 -4x$
b.$g(x)=2x^4 + 5 $
c.$h(x)= \dfrac{x^2-1}{x^2 – 4}$
**Answer Key**

1.
a. Local max:$\left(-\dfrac{1}{\sqrt{2}},2\sqrt{2}\right)$; Local min:$\left(\dfrac{1}{\sqrt{2}},-2\sqrt{2}\right)$
b. Local min:$(0,-1)$
c. No local minimum and maximum
2.
a. No local minimum and maximum
b. Local max:$(0,0)$; Local min:$\left(-\sqrt{\dfrac{3}{2}},-\dfrac{9}{2}\right)$ and $\left(\sqrt{\dfrac{3}{2}},\dfrac{9}{2}\right)$
c. Local min:$\left(0,\dfrac{1}{4}\right)$
*Images/mathematical drawings are created with GeoGebra.*

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