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Chapter 1: Problem 40

Sketch the graph of the function. $$k(x)=\left\\{\begin{array}{ll}2 x+1, & x \leq-1 \\\2 x^{2}-1, & -11\end{array}\right.$$

Short Answer

Expert verified
The sketch of this piecewise function will combine a line going through (1,2) and extending to the left for \(x\leq-1\), a 'squished' and downshifted parabola for \(-11\). It's essential to ensure that the open and closed dots are plotted accurately on the graph, signifying 'less than/greater than' and 'less than or equal to/greater than or equal to' respectively.

Step by step solution

01

Understand the Pieces

The piecewise function \(k(x)\) consists of three 'pieces': \(2x+1\) for \(x \leq -1\), \(2x^{2}-1\) for \(-1 1\). These are a linear function, a quadratic function and another different quadratic function respectively.
02

Sketch the First Piece

The first piece, \(2x+1\), is a linear function with a slope of 2 and y-intercept of 1. Remember, we only need to graph this for \(x \leq -1\). So, graph a line that goes through the point (1,2) and continues to the left, since our upper limit is at \(x = -1\).
03

Sketch the Second Piece

The second piece, \(2x^{2}-1\), is a quadratic function and forms a parabola facing upwards because the coefficient of \(x^{2}\) is positive. This piece should be graphed from \(-1 < x \leq 1\). At \(x = -1\) and \(x = 1\), draw the graph only as an open circle to denote '<' or '>' respectively. At \(x = 1\), as it is '\( \leq \)', sketch a closed dot. The parabola will appear squished or stretched due to the factor 2, and shifted down by 1 due to the '-1'.
04

Sketch the Third Piece

The last piece, \(1-x^{2}\), is again a quadratic function, but will form a parabola facing downwards because the coefficient of \(x^{2}\) is negative. Plot this only for \(x > 1\). The parabola appears to be flipped vertically due to the '-x^{2}' term, and shifted up by 1 due to the '+1'.

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