Problem: Using the knowledge of transformations guess the graph of the following function:

Solution: We start by identifying the basic function. We can disregard the "+ 1" part, since this is just a shift that we can make, and we can also disregard the operations performed with the argument, multiplication by (−2) and shift by π/2. We are left with cos(x), this is our basic function.

So we start with the graph of cosine and first apply the transformations done to the argument. There are two, in the order of calculation they are scaling by (−2) and shift by π/2. The rule for argument is "last to first", so we first shift the graph of the cosine to the left by π/2, then take it and flip it about the y-axis to account for the "-" in (−2), and finally shrink the graph horizontally, towards the y-axis, twice.

The scaling can be sometimes tricky, it might help to keep track of specific points. We pick some prominent point on the graph waiting to be shrinked, then we move it horizontally towards the y-axis so that its x-coordinate gets halved and we get its new position. If we do it with a few prominent points, we should get the shape right; this procedure may also serve as a check of the correctness of the outcome.

For instance, after we shift the graph and then flip it, the "hill" that was originally in the middle moved to the position x = π/2. Therefore, after shrinking twice, it should be at the position π/4. The same hill can be found numerically, it is reached when the argument of cosine is zero and the equation π/2 − 2x = 0 has solution π/4, confirming that the position of this hill in our picture is correct.

Similarly, the first "dip" on the graph to the right from the origin was (after shifting and flipping) at the position x = −π/2, so after shrinking it should be at the position x = −π/4. We check that it fits with numerical solution: The first dip in cosine is at -π, solving π/2 − 2x = −π we get x = −π/4, exactly as in the picture.

Just to be sure, the second positive intercept of cosine with the x-axis is when the argument of cosine is 3π/2, after shift and flip it is at x = −π, so after the shrink it must be at x = −π/2. Again, solving π/2 − 2x = 3π/2 we confirm that this is correct.

If you are not sure, identify more such points and it should help you.

Anyway, now it is time to apply the shift to the value, which is simple, we just move the graph up by 1.

Note: Again, we can check on major features of this graph by comparing it with results that we calculate. For instance, we know that the local maxima of the graph should be at places where cosine is equal to 1. Cosine is 1 when its argument is of the form 2kπ, so we get the equation

π/2 − 2x = 2kπ.

Solving it we see that the hills should be at positions x = π/4 − kπ = π/4 + kπ. (Since k is an arbitrary integer, we can hide the "-" in it, this was not necessary, it just looks better this way.) This checks with our picture.

Similarly we deduce that the "dips" should be at positions x = 3π/4 + kπ and the "middle points", where cosine was zero, should be at positions x = π/2 + kπ. This all fits with our graph as we guessed it, so we are pretty confident that we got it right.

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