Question

For \(F(x)=x^{3}+3 x\), find each value. (a) \(F(1)\) (b) \(F(\sqrt{2})\) (c) \(F\left(\frac{1}{4}\right)\) (d) \(F(1+h)\) (e) \(F(1+h)-F(1)\) (f) \(F(2+h)-F(2)\)

Short answer

(a) 4, (b) 5√2, (c) 49/64, (d) 4 + 6h + 3h^2 + h^3, (e) 6h + 3h^2 + h^3, (f) 15h + 6h^2 + h^3.

Step-by-step solution

Understanding the Function

The given function is \( F(x) = x^3 + 3x \). To find various values, we will substitute different expressions for \( x \) into the function.

Calculating F(1)

Substitute \( x = 1 \) into the function: \[ F(1) = 1^3 + 3(1) = 1 + 3 = 4 \]. Thus, \( F(1) = 4 \).

Calculating F(\sqrt{2})

Substitute \( x = \sqrt{2} \) into the function: \[ F(\sqrt{2}) = (\sqrt{2})^3 + 3(\sqrt{2}) = 2\sqrt{2} + 3\sqrt{2} = 5\sqrt{2} \]. Thus, \( F(\sqrt{2}) = 5\sqrt{2} \).

Calculating F(\frac{1}{4})

Substitute \( x = \frac{1}{4} \) into the function: \[ F\left(\frac{1}{4}\right) = \left(\frac{1}{4}\right)^3 + 3\left(\frac{1}{4}\right) = \frac{1}{64} + \frac{3}{4} = \frac{1}{64} + \frac{48}{64} = \frac{49}{64} \]. Thus, \( F\left(\frac{1}{4}\right) = \frac{49}{64} \).

Calculating F(1 + h)

Substitute \( x = 1 + h \) into the function: \[ F(1 + h) = (1 + h)^3 + 3(1 + h) = (1 + 3h + 3h^2 + h^3) + (3 + 3h) = 4 + 6h + 3h^2 + h^3 \]. Thus, \( F(1 + h) = 4 + 6h + 3h^2 + h^3 \).

Calculating F(1 + h) - F(1)

Using previous results, \( F(1 + h) = 4 + 6h + 3h^2 + h^3 \) and \( F(1) = 4 \). Therefore, \[ F(1 + h) - F(1) = (4 + 6h + 3h^2 + h^3) - 4 = 6h + 3h^2 + h^3 \].

Calculating F(2 + h)

Substitute \( x = 2 + h \) into the function: \[ F(2 + h) = (2 + h)^3 + 3(2 + h) = 8 + 12h + 6h^2 + h^3 + 6 + 3h = 14 + 15h + 6h^2 + h^3 \]. Thus, \( F(2 + h) = 14 + 15h + 6h^2 + h^3 \).

Calculating F(2 + h) - F(2)

Calculate \( F(2) \) first: \[ F(2) = 2^3 + 3(2) = 8 + 6 = 14 \]. Now, \[ F(2 + h) - F(2) = (14 + 15h + 6h^2 + h^3) - 14 = 15h + 6h^2 + h^3 \].

Key concepts

Polynomial Functions

Polynomial functions are fundamental in mathematics and are expressions consisting of variables and coefficients. The variables can have whole number exponents that are non-negative. In simple terms, a polynomial function can be written in the form: \\[ P(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_1x + a_0 \] \where \(a_0, a_1, \ldots, a_n\) are constants (coefficients), \(x\) is the variable, and \(n\) is the degree of the polynomial. For example, in the given function \( F(x) = x^3 + 3x \), we notice it is a polynomial of degree 3 because the highest power of \(x\) is 3. This particular form is called a cubic polynomial because the leading term is \(x^3\). \

• The **degree** of the polynomial function determines its graph's most general shape, and in this case, a cubic polynomial typically has two critical points, which could be a combination of maxima and minima or points of inflection.
• The **coefficients** also tell us how steep our curve might be or how stretched out it is along the axes. Here, the 3 in \(3x\) affects the steepness of the linear part of the curve.

Polynomials like this are very versatile and appear in numerous areas of mathematics and applied sciences.

Substitution Method

The substitution method is a technique used to evaluate functions by plugging in specified values for the variables. It is a straightforward process often used in algebra to find the value of a function for a given input. When using the substitution method, you replace the variable with a specific value, and then perform the mathematical operations to compute the result. \For instance, with the polynomial function \( F(x) = x^3 + 3x \) given in the exercise, to calculate \( F(1) \), we substitute \( x = 1 \) into the function: \

• First, replace \(x\) with 1, so it becomes \((1)^3 + 3 \cdot 1\).
• Next, simplify this to get \(1 + 3 = 4\).

The same technique applies when substituting other values or expressions, such as \(\sqrt{2}\) or \(1 + h\). \Substitution doesn't just help in solving functions; it's also pivotal in various mathematical concepts such as solving equations and calculus, especially when determining limits and derivatives.

Difference of Functions

The concept of the difference of functions involves subtracting one function from another or finding the change in a function's value as its input changes. This approach often helps to measure how much a function varies over a specific interval or under a tiny change in input. \In the given exercise, tasks such as finding \( F(1+h) - F(1) \) or \( F(2+h) - F(2) \) provide practice in evaluating the function's change. To compute \( F(1+h) - F(1) \), you first evaluate \( F(1+h) \) using previously applied substitution methods and then \( F(1) \). Finally, you subtract the latter from the former: \

• Calculate \( F(1+h) = 4 + 6h + 3h^2 + h^3 \).
• Calculate \( F(1) = 4 \).
• Find \( F(1+h) - F(1) = 6h + 3h^2 + h^3 \).

This kind of operation is significant because it reflects the idea of a function's **increment**, and it's closely related to the concept of derivatives in calculus, representing how functions shift as their inputs vary slightly. This tool is invaluable in several mathematical analyses and applications, from engineering problems to financial modeling.