5.5 Exploring Graphs of the Reciprocal Trigonometric...

5.5 Exploring Graphs of the Reciprocal Trigonometric Functions Sketch a graph of both $f(x) = \sin(x)$ and $f(x) = \csc(x)$, and complete the table for $\csc(x)$: Vertical Asymptotes: Local Maximums: Local Minimums: Positive Intervals: Negative Intervals: Intervals of Increase: Intervals of Decrease: Sketch a graph of both $f(x) = \cos(x)$ and $f(x) = \sec(x)$, and complete the table for $\sec(x)$: Vertical Asymptotes: Local Maximums: Local Minimums: Positive Intervals: Negative Intervals: Intervals of Increase: Intervals of Decrease: Sketch a graph of both $f(x) = \tan(x)$ and $f(x) = \cot(x)$, and complete the table for $\cot(x)$: Vertical Asymptotes: Local Minimums: Local Minimums: Positive Intervals: Negative Intervals: Intervals of Increase: Intervals of Decrease:

Transcript

00:02 In the first problem, we are provided with the function f of x which equals to x minus 1 divided by x minus 2 and we are asked to find out the horizontal and the vertical asymptotes.

00:20 So now let us first begin by finding out the vertical asymptotes.

00:28 X is equal to a is said to be a vertical asymptote if limit x tends to a f of x.

00:36 Equals to infinity.

00:38 So here it can be seen that when we take limit x tends to 2 of the function f of x that is x minus 1 over x minus 2 we obtain 2 minus 1 over 2 minus 2 which is 1 over 0 which is equal to infinity.

00:55 So since the condition of vertical asymptote is satisfied we can say that x is equal to 2 is a vertical asymptote of the given function.

01:11 Next, let us find out the horizontal asymptote.

01:18 Y is equal to l is said to be a horizontal asymptote if limit x tends to infinity f of x equals to l.

01:29 So let us consider limit x tends to infinity of x minus 1 the whole divided by x minus 2.

01:37 So now taking x common in the numerator and in the denominator, we have x times.

01:43 1 minus 1 over x the whole divided by x times 1 minus 2 over x x and x get cancelled now let us substitute x is infinity we have 1 minus 1 over infinity which is 0 the whole divided by 1 minus 2 over infinity which is 0 again so we get 1 over 1 which is equal to 1 so therefore we have y is equal to 1 to be a horizontal a simptode.

02:17 So therefore this is the required answer of the first problem.

02:24 In the second problem for the same function f of x equals to x minus 1 over x minus 2 we are asked to find out the intervals where the function is increasing decreasing and we are also asked to check the concavity.

02:44 So now let us begin by finding out the first derivative we make use of the quotient rule here.

02:49 Here.

02:50 So we have x minus 2 times the derivative of the numerator which is 1 minus the numerator which is x minus 1 times the derivative of the denominator which is 1.

03:00 The whole divided by the denominator whole squared.

03:05 So simplifying this we have x minus 2 minus x plus 1 the whole divided by x minus 2 the whole square.

03:14 So here x and negative x would get cancelled and we are left with negative 2 plus 1 which is negative 1.

03:20 So we have negative 1 over x minus 2 the whole squared.

03:25 So now let us equate this to 0...

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