Problem 4 - The Calculus with Analytic Geometry

Question

Given \(G(x)=\sqrt{2 x^{2}+1}\), find: (a) \(G(-2)\) (b) \(G(0)\) (c) \(G\left(\frac{1}{b}\right)\) (d) \(G\left(\frac{4}{7}\right)\) (e) \(G\left(2 x^{2}-1\right)\) (f) \(\frac{G(x+h)-G(x)}{h}, h \neq 0\)

Step-by-Step Solution

Answer

(a) 3, (b) 1, (c) \( \sqrt{\frac{2}{b^2} + 1} \), (d) \( \frac{9}{7} \), (e) \( \sqrt{8x^4 - 8x^2 + 3} \), (f) \( \frac{\sqrt{2x^2 + 4xh + 2h^2 + 1} - \sqrt{2x^2 + 1}}{h} \)

Step 1: Define the function

The given function is defined as: \[ G(x) = \sqrt{2x^2 + 1} \]

Step 2: Calculate \( G(-2) \)

Substitute \( x = -2 \) into the function: \[ G(-2) = \sqrt{2(-2)^2 + 1} = \sqrt{2(4) + 1} = \sqrt{8 + 1} = \sqrt{9} = 3 \]

Step 3: Calculate \( G(0) \)

Substitute \( x = 0 \) into the function: \[ G(0) = \sqrt{2(0)^2 + 1} = \sqrt{0 + 1} = \sqrt{1} = 1 \]

Step 4: Calculate \( G\left(\frac{1}{b}\right) \)

Substitute \( x = \frac{1}{b} \) into the function: \[ G\left(\frac{1}{b}\right) = \sqrt{2 \left(\frac{1}{b}\right)^2 + 1} = \sqrt{2 \left(\frac{1}{b^2}\right) + 1} = \sqrt{\frac{2}{b^2} + 1} \]

Step 5: Calculate \( G\left(\frac{4}{7}\right) \)

Substitute \( x = \frac{4}{7} \) into the function: \[ G\left(\frac{4}{7}\right) = \sqrt{2 \left(\frac{4}{7}\right)^2 + 1} = \sqrt{2 \left(\frac{16}{49}\right) + 1} = \sqrt{\frac{32}{49} + 1} = \sqrt{\frac{32}{49} + \frac{49}{49}} = \sqrt{\frac{81}{49}} = \frac{9}{7} \]

Step 6: Calculate \( G(2x^2 - 1) \)

Substitute \( x = 2x^2 - 1 \) into the function: \[ G(2x^2 - 1) = \sqrt{2(2x^2 - 1)^2 + 1} = \sqrt{2(4x^4 - 4x^2 + 1) + 1} = \sqrt{8x^4 - 8x^2 + 2 + 1} = \sqrt{8x^4 - 8x^2 + 3} \]

Step 7: Calculate the difference quotient

The difference quotient is given by: \[ \frac{G(x + h) - G(x)}{h}, \; h eq 0 \] First find \( G(x + h) \): \[ G(x + h) = \sqrt{2(x + h)^2 + 1} = \sqrt{2(x^2 + 2xh + h^2) + 1} = \sqrt{2x^2 + 4xh + 2h^2 + 1} \] Then compute the difference quotient: \[ \frac{G(x + h) - G(x)}{h} = \frac{\sqrt{2x^2 + 4xh + 2h^2 + 1} - \sqrt{2x^2 + 1}}{h} \]

Key Concepts

Function substitution

Function substitution involves replacing the variable in the given function with another value or expression. This process helps us evaluate the function at specific points or derive new functions based on the substitution made. In the provided exercise, we substitute different values into the function \(G(x) = \sqrt{2x^2 + 1}\) to find the results at those points. For example, substituting \(x = -2\) and calculating \(G(-2)\). Functions change dynamically based on what you substitute. Practice replacing variables with numbers, fractions, or complex expressions to understand the function’s behavior.

Difference quotient

The difference quotient is a fundamental concept in calculus that represents the average rate of change of a function over an interval. It is used to define the derivative of a function. For a function \(G(x)\), the difference quotient is given by:
\[ \frac{G(x + h) - G(x)}{h} \text{ where } h eq 0 \] In our exercise, we start by finding \(G(x+h)\):
\[G(x + h) = \sqrt{2(x + h)^2 + 1} = \sqrt{2(x^2 + 2xh + h^2) + 1} = \sqrt{2x^2 + 4xh + 2h^2 + 1}\text{.} \] Then, we calculate the difference quotient:
\[ \frac{G(x + h) - G(x)}{h} = \frac{\sqrt{2x^2 + 4xh + 2h^2 + 1} - \sqrt{2x^2 + 1}}{h} \] This quotient helps us understand the function's behavior over tiny intervals, forming the basis for derivatives.

Square root functions

Square root functions are functions that involve the square root of a variable or an expression. They follow the form \(f(x) = \sqrt{g(x)}\) where \(g(x)\) is any function of \(x\). These functions are defined only for values of \(x\) where \(g(x) \geq 0\) since the square root of a negative number is not a real number. In the exercise, the provided function is \(G(x) = \sqrt{2x^2 + 1}\). Notice that the argument under the square root, \(2x^2 + 1\), is always non-negative, ensuring that the function is real for all values of \(x\). Understanding square root functions involves being familiar with their domain, range, and general behavior when graphed.

Algebraic manipulation

Algebraic manipulation involves using algebraic rules and operations to simplify expressions, solve equations, or transform and analyze functions. This process is key to solving various parts of the given exercise. For instance, when substituting \(x = \frac{4}{7}\) into \(G(x)\), we simplify:
\[ G\left(\frac{4}{7}\right) = \sqrt{2\left(\frac{4}{7}\right)^2 + 1} = \sqrt{2\left(\frac{16}{49}\right) + 1} = \sqrt{\frac{32}{49} + 1} = \sqrt{\frac{32}{49} + \frac{49}{49}} = \sqrt{\frac{81}{49}} = \frac{9}{7} \] Algebraic manipulation helps in reducing complexities and makes it easier to work with mathematical expressions. Mastering this skill demands practice and understanding of fundamental algebraic principles such as factoring, distribution, and combining like terms.