Video Transcript
Find, if any, the local maximum and
minimum values of 𝑓 of 𝑥 equals 19 sin 𝑥 plus 15 cos 𝑥, together with their
type.
We recall first of all that at the
critical points of a function, the first derivative 𝑓 prime of 𝑥 is equal to
zero. We’ll also need to recall the cycle
that we can use for differentiating sine and cosine. 𝑓 prime of 𝑥 is, therefore, equal
to 19 cos 𝑥 minus 15 sin 𝑥 and we set this equal to zero. To solve, we can first separate the
two terms onto opposite sides of the equation and then divide both sides of the
equation by both cos 𝑥 and 15 to give sin 𝑥 over cos 𝑥 is equal to 19 over
15. At this point, we recall one of our
trigonometric identities: tan 𝜃 is equal to sin 𝜃 over cos 𝜃. So we have tan 𝑥 equals 19 over
15.
To solve, we apply the inverse tan
function. And we must recall at this point
that in order to differentiate trigonometric functions, we must be working with the
angle measured in radians as the key limits used when we first derive the
derivatives from first principles are only true in radians. So when we evaluate 𝑥 on our
calculators, we must make sure we’re working in radians. We find then that 𝑥 is equal to
0.9025 radians. However, tan 𝑥 is a periodic
function with a period of 𝜋. So there are other solutions to
this equation which will correspond to other critical points of the function 𝑓. This means that critical points
will occur at this 𝑥-value that we’ve just found plus or minus integer multiples
of 𝜋. Adding 𝜋 to our value of 0.9025
gives 4.0441 radians. So this will be the second 𝑥-value
at which a local maximum or minimum occurs.
Next, we need to evaluate the
function 𝑓 of 𝑥 at each of the critical points. For our first critical point, when
𝑥 is equal to 0.9025, we get 24.21 to two decimal places. And at our second critical point,
where 𝑥 is equal to 4.0441, we get negative 24.21 to two decimal places. Now, this makes sense because the
sine and cosine functions each have a horizontal line of symmetry on the 𝑥-axis and
therefore so will a sum or difference of the sine and cosine functions, meaning that
the absolute value of the local maximum will be the same as the absolute value of
the local minimum.
Now, finally, we need to apply the
second derivative test to classify these critical points. So let’s make a little bit of
room. We differentiate 𝑓 prime of 𝑥 to
give negative 19 sin 𝑥 minus 15 cos 𝑥. Now we need to evaluate this
function at each of our critical points, but there’s a trick that we can use
here. As we’ve differentiated twice,
we’ve been halfway around our cycle of differentiation, which means that the second
derivative is actually almost identical to the original function. What’s different is that both terms
are negative instead of positive. But if we factor this negative one
out of our expression, we see that in this instance, 𝑓 double prime of 𝑥 is
actually equal to negative 𝑓 of 𝑥.
The reason this is useful is
because we’ve already evaluated 𝑓 of 𝑥 at each of our critical points. So we can use the values we’ve
already found in order to determine the second derivative at our critical
points. At our first critical point, when
𝑥 is equal to 0.9025, 𝑓 of 𝑥 was equal to 24.21. So the second derivative 𝑓 double
prime of 𝑥 will be equal to negative 24.21. As this is less than zero, it tells
us that this critical point will be a local maximum. At our second critical point, the
value of 𝑓 of 𝑥 was negative 24.21. So the value of 𝑓 double prime of
𝑥 will be positive 24.21 and as this is greater than nought, our second critical
point is a local minimum.
Now these are local minimum and
maximum values. But due to the shape of the graph
of 𝑓 of 𝑥, they’re also the absolute minimum and maximum values of the
function. So we can conclude that the local
and absolute minimum value of the function is negative 24.21 and the local and
absolute maximum value of the function is 24.21.
