Problem 5 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 5-10: If \(a\) and \(h\) are real numbers, find (a) \(f(a)\) (b) \(f(-a)\) (c) \(-f(a)\) (d) \(f(a+h)\) (e) \(f(a)+f(h)\) (f) \(\frac{f(a+h)-f(a)}{h}\), if \(h \neq 0\) $$ f(x)=5 x-2 $$

Step-by-Step Solution

Answer

(a) 5a - 2, (b) -5a - 2, (c) -5a + 2, (d) 5a + 5h - 2, (e) 5a + 5h - 4, (f) 5

Step 1: Find f(a)

To find \(f(a)\), substitute \(a\) into the function \(f(x)\). The function is given by \(f(x) = 5x - 2\), so we have:\[f(a) = 5a - 2.\]

Step 2: Find f(-a)

Substitute \(-a\) into the function \(f(x)\). This yields:\[f(-a) = 5(-a) - 2 = -5a - 2.\]

Step 3: Find -f(a)

Take the negative of \(f(a)\), which we found in Step 1:\[-f(a) = -(5a - 2) = -5a + 2.\]

Step 4: Find f(a+h)

Substitute \(a+h\) into the function \(f(x)\):\[f(a+h) = 5(a+h) - 2 = 5a + 5h - 2.\]

Step 5: Find f(a) + f(h)

Calculate \(f(a) + f(h)\) using the function:\[f(h) = 5h - 2,\]so, \[f(a) + f(h) = (5a - 2) + (5h - 2) = 5a + 5h - 4.\]

Step 6: Calculate the Difference Quotient

Calculate \(\frac{f(a+h) - f(a)}{h}\):From Step 4 and Step 1, we have:\[f(a+h) = 5a + 5h - 2,\quad f(a) = 5a - 2. \]Thus,\[f(a+h) - f(a) = (5a + 5h - 2) - (5a - 2) = 5h.\]Now, divide by \(h\):\[\frac{f(a+h) - f(a)}{h} = \frac{5h}{h} = 5. \]

Step 7: Summarize Solutions

For all parts, we find:(a) \(f(a) = 5a - 2\)(b) \(f(-a) = -5a - 2\)(c) \(-f(a) = -5a + 2\)(d) \(f(a+h) = 5a + 5h - 2\)(e) \(f(a) + f(h) = 5a + 5h - 4\)(f) \(\frac{f(a+h) - f(a)}{h} = 5\)

Key Concepts

Difference Quotient

The difference quotient is a powerful tool in calculus and algebra for understanding how functions behave as their input changes. It is specifically defined as\[\frac{f(a+h) - f(a)}{h},\]where \(h\) is not equal to zero. This expression helps in finding the average rate of change of a function over a small interval \([a, a + h]\). It serves as the foundation for the derivative in calculus. To solve the difference quotient, you calculate \(f(a+h)\) and \(f(a)\), find the difference, and divide by \(h\). For instance, given the function \(f(x) = 5x - 2\), the difference quotient simplifies as follows:

• Calculate \(f(a+h)\) to get \(5a + 5h - 2\).
• Subtract \(f(a) = 5a - 2\) from \(f(a+h)\), resulting in \(5h\).
• Divide the result by \(h\), leading to \(\frac{5h}{h} = 5\).

This shows that for the function \(f(x)\), the average rate of change with respect to \(x\) is consistent at 5.

Function Evaluation

Function evaluation involves substituting a specific value into a given function to calculate its outcome. If you have a function such as \(f(x) = 5x - 2\), this process is straightforward:

• For \(f(a)\), replace \(x\) with \(a\) to get \(5a - 2\).
• For \(f(-a)\), replace \(x\) with \(-a\) to obtain \(-5a - 2\).
• To find \(-f(a)\), take the negative of \(f(a)\), which results in \(-5a + 2\).

By performing these substitutions, you can see how the function behaves with different inputs. It's essential to ensure the function is consistent under these operations.

Real Numbers

Real numbers are the set of numbers that include both rational and irrational numbers. They encompass the entire number line. In the context of evaluating functions and the difference quotient, it's crucial to remember that any chosen numbers for \(a\) and \(h\) must be real. Real numbers allow for smooth and continuous changes in functions, which is vital when calculating derivatives or evaluating functions. When performing operations like \(f(a)\) or \(f(a+h)\), the results will be real numbers, provided that the inputs are real numbers as well. This ensures the reliability and applicability of the results across a range of mathematical and real-world problems.

Algebraic Expressions

Algebraic expressions are composed of variables, coefficients, and constants combined using arithmetic operations. They form the basis of the functions we are analyzing:

• For example, \(f(x) = 5x - 2\) is an algebraic expression.
• Expressions like \(5a + 5h - 2\) or \(5a - 2\) represent the output of the function with different inputs.
• Simplification, as seen in the difference quotient \(\frac{5h}{h}\), returns a new algebraic expression, in this case, the constant \(5\).

Understanding how to manipulate and evaluate these expressions is critical for solving algebraic problems, analyzing functions, and applying calculus concepts effectively.