Let F ( X ) = X 3 − 3 X + 14 F(x) = X^3 - 3x + 14 F ( X ) = X 3 − 3 X + 14 . What Is The Local Minimum Value Of F F F Attained At A Critical Point?7) Let F ( X ) = 38 X 2 + 40 6 F(x) = 38x^2 + \frac{40}{6} F ( X ) = 38 X 2 + 6 40 ​ , Where 0 ≤ X ≤ 6 0 \leq X \leq 6 0 ≤ X ≤ 6 . What Is The Estimated Area Obtained By Dividing The

Introduction

In mathematics, finding local minimum values and estimated areas are crucial concepts that have numerous applications in various fields, including physics, engineering, and economics. In this article, we will explore two problems related to these concepts and provide step-by-step solutions.

Problem 1: Finding Local Minimum Values

Problem Statement

Let $f(x) = x^3 - 3x + 14$. What is the local minimum value of $f$ attained at a critical point?

Solution

To find the local minimum value of $f(x)$, we need to find the critical points of the function. A critical point is a point where the derivative of the function is equal to zero or undefined.

First, let's find the derivative of $f(x)$ using the power rule:

$$f'(x) = \frac{d}{dx} (x^3 - 3x + 14) = 3x^2 - 3$$

Now, we need to find the critical points by setting the derivative equal to zero:

$$3x^2 - 3 = 0$$

Solving for $x$, we get:

$$x^2 = 1$$

$$x = \pm 1$$

So, the critical points are $x = 1$ and $x = -1$.

Next, we need to determine which critical point corresponds to a local minimum. We can do this by using the second derivative test. The second derivative of $f(x)$ is:

$$f''(x) = \frac{d^2}{dx^2} (x^3 - 3x + 14) = 6x$$

Evaluating the second derivative at the critical points, we get:

$$f''(1) = 6(1) = 6 > 0$$

$$f''(-1) = 6(-1) = -6 < 0$$

Since $f''(1) > 0$, we know that $x = 1$ corresponds to a local minimum.

Now, we need to find the value of the function at the local minimum:

$$f(1) = (1)^3 - 3(1) + 14 = 1 - 3 + 14 = 12$$

Therefore, the local minimum value of $f(x)$ is $12$.

Problem 2: Finding Estimated Areas

Problem Statement

Let $f(x) = 38x^2 + \frac{40}{6}$, where $0 \leq x \leq 6$. What is the estimated area obtained by dividing the region under the curve into $n$ equal subintervals?

Solution

To find the estimated area, we can use the Riemann sum formula:

$$A \approx \sum_{i=1}^n f(x_i) \Delta x$$

where $x_i$ is the $i$th point in the subinterval, $\Delta x$ is the width of the subinterval, and $n$ is the number of subintervals.

Since we are dividing the region into $n$ equal subintervals, we can write:

$$\Delta x = \frac{b - a}{n} = \frac{6 - 0}{n} = \frac{6}{n}$$

where $a$ and $b$ are the endpoints of the interval.

Now, we need to find the points $x_i$ in each subinterval. We can do this by using the formula:

$$x_i = a + i \Delta x = 0 + i \frac{6}{n} = \frac{6i}{n}$$

where $i$ is an integer ranging from $1$ to $n$.

Substituting the values of $x_i$ and $\Delta x$ into the Riemann sum formula, we get:

$$A \approx \sum_{i=1}^n \left(38 \left(\frac{6i}{n}\right)^2 + \frac{40}{6}\right) \frac{6}{n}$$

Simplifying the expression, we get:

$$A \approx \sum_{i=1}^n \left(\frac{912i^2}{n^2} + \frac{40}{n}\right)$$

To find the estimated area, we need to evaluate the sum for a large value of $n$. Let's use $n = 100$:

$$A \approx \sum_{i=1}^{100} \left(\frac{912i^2}{100^2} + \frac{40}{100}\right)$$

Evaluating the sum, we get:

$$A \approx 123.42$$

Therefore, the estimated area obtained by dividing the region under the curve into $100$ equal subintervals is approximately $123.42$.

Conclusion

Introduction

In our previous article, we explored two problems related to finding local minimum values and estimated areas. In this article, we will answer some frequently asked questions (FAQs) related to these concepts.

Q: What is a local minimum value?

A: A local minimum value is the smallest value of a function in a given interval. It is a point where the function changes from decreasing to increasing.

Q: How do I find the local minimum value of a function?

A: To find the local minimum value of a function, you need to find the critical points of the function. A critical point is a point where the derivative of the function is equal to zero or undefined. You can use the power rule and the second derivative test to find the local minimum value.

Q: What is the second derivative test?

A: The second derivative test is a method used to determine whether a critical point corresponds to a local minimum, maximum, or saddle point. If the second derivative is positive at a critical point, it corresponds to a local minimum. If the second derivative is negative, it corresponds to a local maximum. If the second derivative is zero, it corresponds to a saddle point.

Q: How do I use the Riemann sum formula to estimate the area under a curve?

A: To use the Riemann sum formula to estimate the area under a curve, you need to divide the region under the curve into equal subintervals. You can then use the formula:

$$A \approx \sum_{i=1}^n f(x_i) \Delta x$$

where $x_i$ is the $i$th point in the subinterval, $\Delta x$ is the width of the subinterval, and $n$ is the number of subintervals.

Q: What is the difference between a local minimum value and a global minimum value?

A: A local minimum value is the smallest value of a function in a given interval. A global minimum value is the smallest value of a function over its entire domain.

Q: How do I find the global minimum value of a function?

A: To find the global minimum value of a function, you need to find the critical points of the function and evaluate the function at each critical point. You can then compare the values to find the smallest one.

Q: What is the relationship between the Riemann sum formula and the definite integral?

A: The Riemann sum formula is an approximation of the definite integral. As the number of subintervals increases, the Riemann sum formula approaches the definite integral.

Q: How do I use the definite integral to find the area under a curve?

A: To use the definite integral to find the area under a curve, you need to evaluate the integral:

$$A = \int_{a}^{b} f(x) dx$$

where $a$ and $b$ are the endpoints of the interval.

Conclusion

In this article, we have answered some frequently asked questions related to finding local minimum values and estimated areas. We have provided step-by-step solutions and explanations to help you understand these concepts.