Problem 36 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

Let \(f(x)=2 x^{2}+x\). Find the following. (a) \(f(3)\) (b) \(f(2 x)\) (c) \(f(1+x)\) (d) \(f\left(\frac{1}{x}\right)\) (e) \(\frac{1}{f(x)}\)

Step-by-Step Solution

Answer

(a) \(f(3) = 21\)\n (b) \(f(2x) = 8x^2 + 2x\)\n (c) \(f(1+x) = 2x^2 + 5x + 3\)\n (d) \(f(1/x) = 2/x^2 + 1/x\)\n (e) \(1/f(x) = 1/(x*(2x + 1))\)

Step 1: Substitute x = 3 into the function \(f(x)\)

Substitute 3 for x in the equation \(f(x) = 2x^2 + x\). So, \(f(3) = 2*(3)^2 + 3\). This simplifies to \(f(3) = 18 + 3 = 21\).

Step 2: Substitute x = 2x into the function \(f(x)\)

Substitute 2x for x in the function \(f(x) = 2x^2 + x\). So, \(f(2x) = 2*(2x)^2 + 2x\). This simplifies to \(f(2x) = 8x^2 + 2x\).

Step 3: Substitute x = (1+x) into the function \(f(x)\)

Substitute (1+x) for x in the function \(f(x) = 2x^2 + x\). So, \(f(1+x) = 2*(1+x)^2 + (1+x)\). Expand and simplify to get \(f(1+x) = 2x^2 + 4x + 2 + x + 1 = 2x^2 + 5x + 3\).

Step 4: Substitute x = 1/x into the function \(f(x)\)

Substitute (1/x) for x in the function \(f(x) = 2x^2 + x\). So, \(f(1/x) = 2*(1/x)^2 + 1/x\). Simplify to get \(f(1/x) = 2/x^2 + 1/x\).

Step 5: Find the inverse of the function \(f(x)\)

To find \(1/f(x)\), let's write \(f(x) = 2x^2 + x\) as \(f(x) = x(2x + 1)\). So, \(1/f(x) = 1/(x*(2x + 1))\).

Key Concepts

Polynomial Functions

Polynomial functions are expressions consisting of variables and coefficients, involving only addition, subtraction, multiplication, and non-negative integer exponents of variables. They are one of the most fundamental concepts in algebra and can describe a vast range of mathematical situations. For example, a polynomial function like the one given in the exercise is expressed as:

• The function \(f(x) = 2x^2 + x\) is a quadratic polynomial because the highest power of \(x\) is 2.
• Polynomial functions can be used to model various situations such as projectile paths, area calculations, or any situation with a parabolic relationship.
• The coefficients in the polynomial function (like 2 and 1 in the function \(f(x)\)) determine the shape and position of its graph.

Substitution Method

The substitution method involves replacing the variable in a function with expressions or specific values to evaluate the function. This technique simplifies understanding how changes in the input affect the output.
In our example, the method is used to find values like \(f(3)\) or \(f(2x)\), where:

• For \(f(3)\), we substitute \(3\) into every \(x\) in the function, leading to \(f(3) = 2 \times (3)^2 + 3\).
• For \(f(2x)\), the entire expression \(2x\) is substituted for every \(x\) in the function, resulting in recalculating to form \(f(2x) = 8x^2 + 2x\).

This method is essential for evaluating functions at specific points or using algebraic transformations for broader applications.

Inverse Function

An inverse function essentially "undoes" the work of the original function. However, this concept slightly alters when talking about reciprocal functions, like \(1/f(x)\). Here, we are finding a function that gives us the reciprocal of the original function's output rather than its inverse.
The exercise asks for \(1/f(x)\), calculated as:

• First, express \(f(x)\) in a simpler product form: \(f(x) = x(2x+1)\).
• Then, the reciprocal \(1/f(x)\) follows as \(1/(x(2x+1))\).

It's helpful because reciprocal functions can reveal behaviors in systems where inverse relations are considered, like in fractions or rates.

Quadratic Functions

Quadratic functions are a type of polynomial function where the highest exponent of the variable is 2, forming a parabolic graph when visualized. Our specific example, \(f(x) = 2x^2 + x\), is a quadratic function.
Important features of quadratic functions include:

• The standard form: \( ax^2 + bx + c\). Here, \(a = 2\), \(b = 1\), and \(c = 0\).
• The parabola's vertex, representing the function's minimum or maximum value depending on the coefficient \(a\).
• The direction of the parabola (upwards or downwards) is determined by the sign of \(a\). Since \(a = 2\) is positive, the parabola opens upwards.
• The axis of symmetry is located at \(x = -\frac{b}{2a}\), helping to understand the symmetry of the function's graph.
• Solutions to the quadratic function, or "roots," can be found using the quadratic formula when set equal to zero.

Quadratic functions model numerous physical phenomena, including motion and optimization problems.