Use the second derivative test to find the relative maximum and minimum of the function. f(t) =...

Question:

Use the second derivative test to find the relative maximum and minimum of the function.

{eq}\displaystyle\; f(t) = \frac{\left(t + 3\right)^{3}}{\left(t - 2\right)^{2}} {/eq}

Second Derivative Test:

The second derivative "test" is used to identify and classify extrema. We use parenthesis because of the negative connotations associated with this word, plus all we are doing is taking advantage of what the second derivative actually is; we are not "testing" anything, we are only applying concepts. Recall that the first derivative gives us the slope of the function, so extrema are likely to occur where it equals 0. And the second derivative tells us about the concavity of the curve: upward concavity means a curve is bowl shaped, downward concavity means a curve is shaped like an upside down bowl.

Answer and Explanation:

We need the first two derivatives of the function. We have

{eq}\begin{align*} f' (t) &= \frac{d}{dt} \left( \frac{\left(t + 3\right)^{3}}{\left(t - 2\right)^{2}} \right) \\ &= \frac{3(t+3)^2 \cdot (t-2)^2 - \left(t + 3\right)^{3} \cdot 2(t-2)}{\left(t - 2\right)^{4}} \\ &= \frac{(t+3)^2(t-2)(t-12)}{\left(t - 2\right)^{4}} \\ &= \frac{(t+3)^2(t-12)}{\left(t - 2\right)^{3}} \end{align*} {/eq}

Note that the zeroes are at {eq}t = -3,\ 12 {/eq}, and there is a vertical asymptote at {eq}t = 2 {/eq}. Now, we need the second derivative to classify these potential extrema (all but {eq}t = 2 {/eq}). We differentiate again to find

{eq}\begin{align*} f'' (t) &= \frac{d}{dt} \left( \frac{(t+3)^2(t-12)}{\left(t - 2\right)^{3}} \right) \\ &= \frac{\left[ 2(t+3)(t-12) + (t+3)^2(1) \right] \cdot (t-2)^3 - (t+3)^2(t-12) \cdot 3(t-2)^2}{\left(t - 2\right)^{6}} \\ &= \frac{150(t-2)^2(t+3)}{\left(t - 2\right)^{6}} \\ &= \frac{150(t+3)}{\left(t - 2\right)^{4}} \end{align*} {/eq}

Note that when {eq}t = -3 {/eq}, the second derivative is 0, and so it is an inflection point. Then when {eq}t < -3 {/eq}, the second derivative is negative, which means the curve is concave down here. This means that any potential extrema to the left of {eq}t = -3 {/eq} must be a relative maxima (since the curve is shaped like an upside down bowl here). Also, when {eq}t > -3 {/eq}, the second derivative is positive, which means we have positive concavity, and so we have minima to the right of the inflection point. This means that...

...at {eq}t = -3 {/eq}, we have an inflection point; this is not an extremum.

...at {eq}t= 12 {/eq}, we have a relative minimum of

{eq}\begin{align*} f(12) &= \frac{\left(12 + 3\right)^{3}}{\left(12 - 2\right)^{2}} \\ &= \frac{135}4 \\ &= 33.75 \end{align*} {/eq}

Note that nowhere along the way did we "test" anything. We simply got what we got and then we thought about what it must mean. The best thing about mathematics is that it really is just applied logic, so as long as we understand what's going on, we rarely need to memorize anything.


Learn more about this topic:

Loading...
Concavity and Inflection Points on Graphs

from Math 104: Calculus

Chapter 9 / Lesson 5
7.1K

Related to this Question

Explore our homework questions and answers library