Problem 12 - Single Variable Calculus

Question

In Exercises \(7-24\), sketch the graph of the function and find its absolute maximum and absolute minimum values, if any. $$ h(x)=x^{2}+1 \text { on }(-2,1] $$

Step-by-Step Solution

Answer

The absolute minimum value of the function \(h(x) = x^2 + 1\) on the interval \((-2, 1]\) is 1 at \(x=0\), and the absolute maximum value is 5 at \(x=-2\).

Step 1: Identify the critical points

To find the critical points, we need to take the derivative of the function \(\displaystyle h(x) \), and then set it equal to 0, to find the x-coordinates of any critical points. Let's find the derivative \(\displaystyle h'(x) \) : $$ h'(x) = \frac{d}{dx} (x^2 + 1) = 2x $$ Now, set the derivative to 0 and solve for \(\displaystyle x\): $$ 2x = 0 \Rightarrow x = 0 $$ So, the critical point is \(\displaystyle x = 0\).

Step 2: Evaluate the function at the critical points and endpoints

To find the absolute maximum and minimum values, we will evaluate the function \(\displaystyle h(x) \) at the critical point and at the endpoints of the interval \((-2, 1]\). We have: $$ h(0) = (0)^2 + 1 = 1 $$ For the endpoints, we have: $$ h(-2) = (-2)^2 + 1 = 4 + 1 = 5 $$ Note that \(\displaystyle h(1) \) is already within the interval and this case was already considered as a critical point. Thus, we don't need to calculate it again.

Step 3: Identify the maximum and minimum values

Now, compare the function values calculated in the previous step: $$ h(0) = 1\\ h(-2) = 5 $$ As we can see, the minimum value of the function on the interval \((-2, 1]\) is 1, which occurs at \(\displaystyle x=0\), and the maximum value of the function is 5, which occurs at \(\displaystyle x=-2\). #Conclusion# The graph of the function \(h(x) = x^2 + 1\) on the interval \((-2, 1]\) has an absolute minimum value of 1 at \(\displaystyle x=0\) and an absolute maximum value of 5 at \(\displaystyle x=-2\).

Key Concepts

Absolute Maximum and Minimum

In calculus, the absolute maximum and minimum refer to the highest and lowest values a function attains over a given interval. To determine these, you need to evaluate the function at all critical points and endpoints of the interval. This provides a full picture of the function's behavior.

For the function in our exercise,

• the critical point was found by setting the derivative equal to zero, which indicated potential maximum or minimum points by the stationary condition of the graph.
• The endpoints of the interval also play a significant role, as potential extremal points.

By evaluating the function at both these critical points and endpoints, we identify the actual absolute maximum and minimum numbers. In this problem:

• The absolute maximum value is 5 at the endpoint \( x = -2 \).This extreme happens because it's the highest point over the interval.
• The absolute minimum value is 1 at the location of the critical point \( x = 0 \).This is lower than any function value achieved in the interval.

Critical Points

Critical points are specific points in the domain of a function where the derivative equals zero or the derivative does not exist. These points hold significant importance as they indicate where the graph of the function might change direction, which could lead to local maxima or minima.

To identify critical points,

• First find the derivative of the function, as these points are closely related to the behavior of the derivative.
• Then solve the equation \( h'(x) = 0 \).

In our case, solving the derivative \( h'(x) = 2x = 0 \) gives a critical point at \( x = 0 \). At this point, the slope of the tangent to the curve is zero, indicating a potential extremum. Evaluating the function at this point helps determine if it's an absolute or relative extremum.

Derivative

The derivative of a function indicates how the function's value changes as its input changes. It's essentially a rate of change or a slope of the tangent line to the curve at a point. Calculating the derivative is crucial for identifying the behaviors of the graph, such as where slopes are increasing, decreasing, or stationary.

For our function \( h(x) = x^2 + 1 \),

• The derivative is \( h'(x) = 2x \), which was determined by differentiating the original function.
• Setting \( h'(x) = 0 \) and solving gives the location of potential critical points.

Knowing the derivative allows us to effectively analyze and sketch the graph of the function, as it describes how each input value affects the output within its domain.

Graph of a Function

The graph of a function gives a visual representation of all the points \( (x, h(x)) \), showing how the output value changes over the domain as the input changes. For the function \( h(x) = x^2 + 1 \), the graph is a parabola shifted one unit up from the standard \( x^2 \) parabola.

Specific features on the graph include:

• Vertex: The lowest point of the parabola, located at \( x = 0 \), is the absolute minimum in our interval.
• Direction: Since the parabola opens upward, it indicates increasing values as \( x \) moves away from the vertex.
• End behavior within the interval \( (-2, 1] \): As the endpoint \( x = -2 \) is approached, the graph shows maximum output.

Understanding the graph fully allows us to confirm calculated values for absolute maximums and minimums visually and is vital for connecting algebraic solutions with visual insights.