/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Draw the graphs of $$ f(x)=\... [FREE SOLUTION] | 91影视

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Chapter 2: Problem 24

Draw the graphs of $$ f(x)=\sin ^{2} x-1 $$

Short Answer

Expert verified
The graph of \(f(x)=\sin ^{2} x-1\) oscillates between -1 and 0. It reaches its maximum value of 0 at \( \pi/2, 3 \pi/2 \) and so on and reaches its minimum value of -1 at \( 0, \pi, 2\pi \) and so on. The function has a period of \( 2\pi \).

Step by step solution

01

Identify the Basic Function

The base function here is \( \sin x \). The square of this function \( \sin^{2}x \) oscillates between 0 and 1 in its basic period \(0 \leq x \leq 2\pi \), and this will be our starting point.
02

Apply the Transformation

Note that we have subtracted 1 from \( \sin^{2}x \). This will shift all values of the original function down by one. Therefore, the values of the function are shifted to range between -1 and 0.
03

Sketch the Graph

To sketch the graph, plot key points such as \( x = 0, \pi/2, \pi, 3\pi/2, \) and \( 2\pi \). You will see that the graph starts at -1 at x = 0, goes up to 0 at \( \pi/2 \), returns to -1 at \( \pi \), goes up to 0 at \( 3\pi/2 \) and finally returns to -1 at \( 2\pi \). Do notice that the function is periodic with period 2\pi

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