Problem 3 - The Calculus with Analytic Geometry

Question

Given \(F(x)=\sqrt{2 x+3}\), find: (a) \(F(-1)\) (b) \(F(4)\) (c) \(F\left(\frac{t}{2}\right)\) (d) \(F(30)\) (e) \(F(2 x+3)\) (f) \(\frac{F(x+h)-F(x)}{h}, h \neq 0\)

Step-by-Step Solution

Answer

a) 1, b) \sqrt{11}\, c) \sqrt{t + 3}\, d) \sqrt{63}\, e) \sqrt{4x + 9}\, f) \frac{\sqrt{2x + 2h + 3} - \sqrt{2x + 3}}{h}\.

Step 1: Evaluate at the given point for (a)

To find \(F(-1)\), substitute \x = -1\ in the function \F(x) = \sqrt{2x + 3}\. \F(-1) = \sqrt{2(-1) + 3} = \sqrt{-2 + 3} = \sqrt{1} = 1\.

Step 2: Evaluate at the given point for (b)

To find \(F(4)\), substitute \x = 4\ in the function \F(x) = \sqrt{2x + 3}\. \F(4) = \sqrt{2(4) + 3} = \sqrt{8 + 3} = \sqrt{11}\.

Step 3: Evaluate at the given expression for (c)

To find \(F\bigg(\frac{t}{2}\bigg)\), substitute \x = \frac{t}{2}\ in the function \F(x) = \sqrt{2x+3}\. \F\bigg(\frac{t}{2}\bigg) = \sqrt{2\bigg(\frac{t}{2}\bigg) + 3} = \sqrt{t + 3}\.

Step 4: Evaluate at the given point for (d)

To find \(F(30)\), substitute \x = 30\ in the function \F(x) = \sqrt{2x + 3}\. \F(30) = \sqrt{2(30) + 3} = \sqrt{60 + 3} = \sqrt{63}\.

Step 5: Evaluate at the given expression for (e)

To find \(F(2x+3)\), substitute \x = 2x + 3\ in the function \F(x) = \sqrt{2x+3}\. \F(2x+3) = \sqrt{2(2x+3) + 3} = \sqrt{4x + 6 + 3} = \sqrt{4x + 9}\.

Step 6: Evaluate the difference quotient for (f)

To find \frac{F(x+h) - F(x)}{h} \, compute \(F(x+h)\) first. \F(x + h) = \sqrt{2(x + h) + 3} = \sqrt{2x + 2h + 3}\. Then find the difference quotient: \frac{F(x+h) - F(x)}{h} = \frac{\sqrt{2x + 2h + 3} - \sqrt{2x + 3}}{h}\.

Key Concepts

Calculus

Calculus is a branch of mathematics that focuses on change and motion. It has two main parts: differential calculus and integral calculus. Differential calculus deals with the concept of a derivative, which represents how a function changes as its input changes. On the other hand, integral calculus deals with the concept of an integral, which represents the accumulation of quantities. Calculus is foundational in many fields, including physics, engineering, economics, and medicine. In the context of function evaluation, calculus helps us understand how a function behaves as we tweak different variables.

Difference Quotient

The difference quotient is an expression used in calculus to find the slope of the secant line between two points on the graph of a function. It's given by the formula \(\frac{F(x+h) - F(x)}{h}\), where \(F(x)\) is the function, \(x\) is a point, and \(h\) is a small step from \(x\). This quotient is crucial because it lays the groundwork for the derivative, which represents the slope of the tangent line at a specific point. Let's apply it to our function \(F(x) = \sqrt{2x + 3}\). By substituting \(x+h\), we get:

\t
• \(F(x+h) = \sqrt{2(x + h) + 3} = \sqrt{2x + 2h + 3}\)

Next, we form the difference quotient:

\t
• \(\frac{F(x+h) - F(x)}{h} = \frac{\sqrt{2x + 2h + 3} - \sqrt{2x + 3}}{h}\)

This quotient measures the average rate of change of the function over the interval \([x, x+h]\). As \(h\) approaches zero, this quotient approaches the derivative of the function at \(x\).

Square Root Function

A square root function is a type of function that includes a square root of an expression involving the variable. Its general form is \(F(x) = \sqrt{expression}\), like \(F(x) = \sqrt{2x + 3}\). The function \(F(x)\) maps a number \(x\) to the \(\sqrt{2x + 3}\). It is defined as long as the expression inside the square root is non-negative. For \(F(x) = \sqrt{2x + 3}\), the domain of the function is \(x \geq -1.5\) since \(2x + 3\) should be \(\geq 0\).

When evaluating the square root function at different points such as \(F(-1)\), \(F(4)\), \(F(\frac{t}{2})\), and so on, always make sure to first plug in the value or expression in place of \(x\). Then, simplify the expression inside the square root before taking the square root.

Substitution

Substitution is a simple but powerful tool in algebra and calculus. It involves replacing a variable with a specific value or another expression. This technique is frequently used to evaluate functions at specific points. For example, if we want to evaluate \(F(4)\) for the function \(F(x) = \sqrt{2x + 3}\), we substitute \(4\) for \(x\) and perform the arithmetic:

\t
• \(F(4) = \sqrt{2(4) + 3} = \sqrt{8 + 3} = \sqrt{11}\)

Similarly, by substituting a more complex expression like \(\frac{t}{2}\) or \(2x + 3\), it’s essential to follow the same steps:

\t
• First, replace \(x\) with the given expression.
\t
• Simplify the expression inside the function.
\t
• Finally, apply the function’s rule if needed.

Substitution is also used in the difference quotient, where we substitute \(x + h\) for \(x\). Understanding substitution well makes calculus and algebra problem-solving much easier and systematic.