Question

\(y=\left\\{\begin{array}{ll}{-\frac{1}{4} x^{2}-\frac{1}{2} x+\frac{15}{4},} & {x \leq 1} \\ {x^{3}-6 x^{2}+8 x,} & {x>1}\end{array}\right.\)

Short answer

This is a piecewise function, with two parts. First part is a quadratic function \(-\frac{1}{4} x^{2}-\frac{1}{2} x+\frac{15}{4}\) defined for \(x \leq 1\), and the second part is a cubic function \(x^{3}-6 x^{2}+8 x\) defined for \(x > 1\). The value of the function at \(x = 1\) is 3.

Step-by-step solution

Analyze the first piece of the function

The first piece of the function is \(-\frac{1}{4} x^{2}-\frac{1}{2} x+\frac{15}{4}\), and it is defined for \(x \leq 1\). Handle it as any other quadratic function.

Analyze the second piece of the function

The second piece of the function is \(x^{3}-6 x^{2}+8 x\), and it is defined for \(x > 1\). This is a cubic function.

Understand domain of definition for each part

The first part is defined for \(x \leq 1\) while the second part is defined for \(x > 1\). It means that the value of \(x = 1\) will be served by the first part of the function (-\frac{1}{4}*1^{2}-\frac{1}{2}*1+\frac{15}{4} = 3).

Key concepts

Quadratic Functions

A quadratic function is a type of polynomial that can be represented by the standard form equation: \( y = ax^2 + bx + c \), where \( a \), \( b \), and \( c \), are coefficients, with \( a \) not equal to zero. The graph of a quadratic function is a smooth, symmetrical curve called a parabola that opens upwards or downwards depending on the sign of \( a \). The highest or lowest point of a parabola is known as the vertex. In the given exercise, the quadratic function is \( y = -\frac{1}{4}x^2 - \frac{1}{2}x + \frac{15}{4} \), and its graph will open downwards since the coefficient of \( x^2 \) is negative.

The domain of a quadratic function is all real numbers, \( x \in \mathbb{R} \), since no matter what x-value we choose, you can always calculate a corresponding y-value. However, textbook exercises will often specify a subdomain where the function applies, tailoring it to a specific context or merging it with other functions to create piecewise functions. In the exercise's first part, the quadratic function's domain is limited to \( x \leq 1 \).

Cubic Functions

Whereas quadratic functions involve the square of the variable, cubic functions feature variables raised to the power of three. The general form of a cubic function is \( y = ax^3 + bx^2 + cx + d \), where \( a \) is nonzero. Cubic functions can occasionally look like wiggly lines that pass through the graph multiple times. They have unique characteristics, such as having at most two turning points and possibly a point of inflection where the function changes concavity.

In our exercise, the cubic function is given by \( y = x^3 - 6x^2 + 8x \). It exhibits the dynamic behavior typical of cubic functions, with changes in direction that imply a more complex relationship between x and y compared to quadratic functions. For the second piece of the function in the problem, the domain is specified as \( x > 1 \), highlighting that this function 'takes over' from the quadratic piece for values of \( x \) greater than 1.

Domain of Definition

The domain of definition refers to the set of all possible input values, typically represented by the variable x, for which a function is defined. Understanding the domain is crucial because it tells us the values we can plug into a function to get a valid output. For continuous functions, the domain is usually all real numbers, but it can also be limited by specific constraints or by the nature of the function.

Piecewise functions like the one in our example have different rules for different intervals of the domain. It's essential to identify which part of the function applies to which interval of x-values. The first part of the problem's function, a quadratic, is applicable for \( x \leq 1 \), while the second part, a cubic function, is applicable for \( x > 1 \). You should always consider the domain explicitly stated in a piecewise function to understand how the function behaves based on the input value of x.