Question

Critical points and extreme values a. Find the critical points of the following functions on the given interval. Use a root finder, if necessary. b. Use a graphing utility to determine whether the critical points correspond to local maxima, local minima, or neither. E. Find the absolute maximum and minimum values on the given inter. wal, if they exist. $$g(x)=(x-3)^{5 / 3}(x+2) \text { on }[-4,4]$$

Short answer

Question: Find the critical points of the function $$g(x) = (x-3)^{5/3}(x+2)$$ on the interval [-4, 4], determine their nature (local maxima, local minima, or neither), and find the absolute maximum and minimum values on the interval. Solution: The critical points are $$x = 3$$ and $$x = -\frac{1}{8}$$. At $$x=3$$, there is a local minimum, and at $$x = -\frac{1}{8}$$, there is a local maximum. The absolute maximum value is $$g(-\frac{1}{8}) = (\frac{23}{8})^{\frac{5}{3}}(\frac{15}{8})$$, and the absolute minimum value is $$g(3) = 0$$.

Step-by-step solution

Find the critical points

To find the critical points, we must first find the derivative of $$g(x)$$. We can use the product rule: $$(uv)' = u'v + uv'$$. Let $$u = (x-3)^{5/3}$$ and $$v = (x+2)$$. We need to find $$u'$$ and $$v'$$: $$u' = \frac{5}{3}(x-3)^{2/3}$$ $$v' = 1$$ Now, we can find $$g'(x)$$ using the product rule: $$g'(x) = u'v + uv' = (x-3)^{5/3}(1) + (x+2)\left(\frac{5}{3}(x-3)^{2/3}\right)$$ $$g'(x) = (x-3)^{5/3} + (x+2)\left(\frac{5}{3}(x-3)^{2/3}\right)$$ Critical points occur when $$g'(x) = 0$$ or $$g'(x)$$ is undefined. Let's find these points: $$0 = (x-3)^{5/3} + (x+2)\left(\frac{5}{3}(x-3)^{2/3}\right)$$ Solving this equation might be a bit tricky yet, so let's focus on the undefined points first. $$g'(x)$$ would be undefined if the denominator of the fractional part is zero, but there is no such fractional part in our case. Now let's return to solving $$g'(x) = 0$$: Factor out $$(x-3)^{2/3}$$: $$(x-3)^{2/3}((x-3)+(x+2)\frac{5}{3}) = 0$$ Now we have two possibilities: 1) $$(x-3)^{2/3} = 0 \implies x = 3$$, which is within our interval. 2) $$(x-3) + (x+2)\frac{5}{3} = 0$$. Solve for x: $$x + \frac{5}{3}(x+2) = 3$$ $$x + \frac{5x}{3} + \frac{10}{3} = 3$$ $$\frac{8x}{3} = -\frac{1}{3}$$ $$x = -\frac{1}{8},$$ which is also within our interval. So, our critical points are $$x = 3$$ and $$x = -\frac{1}{8}$$.

Determine the nature of the critical points

Now we can analyze the behavior of the function near the critical points using a graphing utility, which will allow us to determine if they are maxima, minima, or neither. By graphing the function, we can see that at $$x=3$$ the function has a local minimum, and at $$x = -\frac{1}{8}$$ the function has a local maximum.

Find the absolute maximum and minimum values

We need to determine the values of the function at our critical points and endpoints. Then we can compare the values to determine the absolute maximum and minimum. Plug x values into the original function: $$g(-4) = (1)(-2) = -2$$ $$g(-\frac{1}{8}) = (\frac{23}{8})^{\frac{5}{3}}(\frac{15}{8})$$ $$g(3) = (0)(5) = 0$$ $$g(4) = (1)(6) = 6$$ So, the absolute maximum value is $$g(-\frac{1}{8}) = (\frac{23}{8})^{\frac{5}{3}}(\frac{15}{8})$$, and the absolute minimum value is $$g(3) = 0$$.

Key concepts

Product Rule Differentiation

Understanding the product rule of differentiation is crucial for solving calculus problems involving the derivative of a function that is the product of two sub-functions. To apply the product rule, we first identify two functions within the product, which we can denote as u and v.

According to the product rule, the derivative of a composite function, uv, will be u'v + uv', where u' and v' are the derivatives of u and v respectively. In our case, we have u = (x-3)^{5/3} and v = (x+2), and their derivatives are u' = \(\frac{5}{3}(x-3)^{2/3}\) and v' = 1.

By following this rule, we ensure that we accurately find the derivative of the product of two functions without omitting any components that are critical to the solution. Failing to apply the product rule or applying it incorrectly can lead to significant errors in finding critical points, which are necessary for determining the behavior of the function regarding maxima and minima.

Absolute Extrema

In calculus, identifying the absolute extrema of a function over a specific interval involves finding the greatest (absolute maximum) and the least (absolute minimum) values that the function assumes. These values describe the overall highest and lowest points on the graph of the function, within the given range.

It's important to remember that absolute extrema can occur at critical points—where the derivative is zero or undefined—or at the endpoints of the interval. To methodically find these extrema, we follow three key steps: first, find the critical points of the function; second, evaluate the function at critical points and the interval's endpoints; and lastly, compare these values to determine which is the highest and which is the lowest.

By employing these steps as we did in our problem, we can assure that the absolute maximum and minimum have been thoroughly examined, ensuring a comprehensive understanding of the function's complete behavior on the interval.

Graphing Utility Analysis

Utilizing graphing utilities, like a graphing calculator or software, can be incredibly helpful for visually analyzing the behavior of a function. These tools go hand-in-hand with calculus concepts as they aid in identifying the type of extremum at a critical point—whether it's a local maximum or minimum, or alternatively, a point of inflection.

In the context of our exercise, after finding the critical points algebraically, we can plot the function using such a utility. By examining the curvature around the critical points, we observe where the function peaks (maxima), where it dips (minima), or if it flattens out without turning (neither). These utilities enhance our comprehension by providing a graphical representation, offering visual confirmation of calculations and revealing the function's nature at critical values.

Derivative of a Function

The derivative of a function is a core concept in calculus, which represents the rate at which the function’s output value changes with respect to changes in the input value. It gives us important information about the behavior of the function, such as slopes of tangent lines, rates of change, and the identification of critical points that are pertinent to our investigation of extrema.

Computing a derivative often requires applying various rules and techniques—such as the product rule, which we’ve seen in the earlier section. It's important to note that the derivative may not always exist for every point in the function's domain; points where it is undefined or does not exist may also provide essential insights into the function's structure.

Finding the derivative is an initial and crucial step in solving many types of calculus problems, including those that involve optimization and motion. Through the lens of the derivative, we're able to uncover a more nuanced and precise understanding of a function's behavior.