Problem 44 - Calculus and its applications

Question

Find \(f^{\prime}(x)\). $$ f(x)=-0.01 x^{2}-0.5 x+70 $$

Step-by-Step Solution

Answer

The derivative of the function is \( f'(x) = -0.02x - 0.5 \).

Step 1: Identify the Function and Its Terms

The function given is \( f(x) = -0.01x^2 - 0.5x + 70 \). This function is a polynomial of degree 2, which has three terms: \(-0.01x^2\) (quadratic), \(-0.5x\) (linear), and \(70\) (constant).

Step 2: Determine Derivative Rules

To find \( f'(x) \), we need to differentiate each term of the function separately. For a function of the form \( ax^n \), the derivative is \( (an)x^{n-1} \). A constant term's derivative is zero.

Step 3: Differentiate the Quadratic Term

Consider the quadratic term \(-0.01x^2\). Using the differentiation rule \( nx^{n-1} \), the derivative is \(-0.02x\) since multiplying the power 2 by the coefficient \(-0.01\) gives us \(-0.02\), and the power of \(x\) decreases by 1.

Step 4: Differentiate the Linear Term

The linear term is \(-0.5x\). The derivative of \(x\) is 1, so multiplying this by the coefficient \(-0.5\) gives us \(-0.5\).

Step 5: Differentiate the Constant Term

The constant term is \(70\). The derivative of a constant is 0 because it doesn't change as \(x\) changes.

Step 6: Combine the Derivatives

Combine the derivatives found from each term: \(-0.02x\) from the quadratic term, \(-0.5\) from the linear term, and \(0\) from the constant term. Thus, the derivative is \( f'(x) = -0.02x - 0.5 \).

Key Concepts

Polynomial Functions

A polynomial function is an expression made up of variables and coefficients, involving operations of addition, subtraction, multiplication, and non-negative integer exponents of variables. In our original exercise, the given function is a second-degree polynomial, also known as a quadratic, which takes the standard form:

• Quadratic term: This is represented by \(-0.01x^2\) in the function, where the variable \(x\) is squared.
• Linear term: This term, \-0.5x\, involves \(x\) raised only to the first power.
• Constant term: The number 70, which stands alone and does not change as \(x\) changes.

Understanding the structure of polynomial functions helps us recognize the impact each term has on the overall shape and behavior of the graph. Each term contributes differently, with quadratic terms curving the graph and linear terms creating slants.

Differentiation Rules

Differentiation is a key process in calculus, which involves finding the derivative of a function. A derivative measures how a function's value changes as its input changes. When differentiating polynomial functions, we use simple rules that follow predictable patterns:

• Power Rule: For a term like \(ax^n\), the derivative is \((an)x^{n-1}\). Essentially, multiply the exponent by the coefficient, and decrease the exponent by one.
• Constant Rule: The derivative of a constant is zero. Constants do not change, so their rate of change is zero.

Applying these rules allows us to differentiate terms easily, as seen in the solution for \(f(x) = -0.01x^2 - 0.5x + 70\). Understanding these rules simplifies many calculus problems and builds a foundation for more advanced concepts.

Calculus Problems

Calculus problems often involve finding derivatives, which describe how things change. By practicing derivative calculations, you gain insights into the rate at which quantities change. This is crucial for understanding motion, growth, and other variable rates in real-world contexts. Solving calculus problems:

• Identify the Function: Recognize what type of function you have. Is it a polynomial, exponential, or another type?
• Use Differentiation Rules: Apply rules like the power rule for polynomial terms to differentiate each part systematically.
• Combine Results: Once each term is differentiated, sum them up to find the derivative of the entire function.

By approaching calculus problems in a structured way, you will find them less daunting and more manageable. These techniques are building blocks for tackling complex scenarios in science and engineering.