Chapter 4 Review: Class Notes
Welcome to the Chapter 4 review! This chapter focuses on key calculus concepts such as finding critical points, determining concavity, and solving optimization problems. Let's dive into some examples to solidify your understanding. Remember, practice makes perfect, and you've got this!
1. Concavity and Critical Points
Let's start with an example of how to determine concavity and find critical points for a given function. Consider the function:
$$y = x^3 + 3x^2$$First, find the first and second derivatives:
$$y' = 3x^2 + 6x$$ $$y'' = 6x + 6$$To find the critical points, set the second derivative equal to zero:
$$6x + 6 = 0$$ $$x = -1$$Now, analyze the concavity:
- For $x < -1$, $y'' < 0$, so the function is concave down.
- For $x > -1$, $y'' > 0$, so the function is concave up.
2. More Complex Concavity Example
Let's tackle a slightly more complex example:
$$y = \frac{x - x^2}{2 - 3x + x^2} = \frac{-x}{x-2}$$First, find the first derivative using the quotient rule:
$$y' = \frac{2}{(x-2)^2}$$Then find the second derivative, again using the quotient rule:
$$y'' = \frac{-4}{(x-2)^3}$$Set the second derivative equal to zero to find any possible inflection points; however, in this case, the second derivative is never zero, but is undefined at $x = 2$. We must consider the concavity on either side of this value:
- If $x < 2$, then $y'' > 0$, so the function is concave up.
- If $x > 2$, then $y'' < 0$, so the function is concave down.
3. Finding Minima and Maxima
Consider the function:
$$y = (x - 4)\sqrt[3]{x} = x^{\frac{4}{3}} - 4x^{\frac{1}{3}}$$First find the derivative:
$$y' = \frac{4}{3}x^{\frac{1}{3}} - \frac{4}{3}x^{-\frac{2}{3}} = \frac{4x-4}{3x^{\frac{2}{3}}}$$Set the derivative equal to zero to find the critical numbers:
$$4x - 4 = 0 \implies x = 1$$ $$3x^{\frac{2}{3}} = 0 \implies x = 0$$Analyzing the sign of $y'$ around these critical numbers, we have a local minimum at $x=1$ and neither a minimum or maximum at $x=0$.
4. Trigonometric Functions: Min/Max
Consider the function:
$$y = \sin^3(x)$$Find the first derivative:
$$y' = 3\sin^2(x)\cos(x)$$Set the derivative equal to zero to find the critical numbers:
$$\sin(x) = 0 \implies x = 0, \pi$$ $$\cos(x) = 0 \implies x = \frac{\pi}{2}, \frac{3\pi}{2}$$By analyzing the sign of the first derivative around these points, we can determine the local maxima and minima on the interval $[0, 2\pi]$.
Keep practicing these concepts, and you'll be well-prepared for Chapter 4! Good luck with your review and test!