Question

Consider the equation below. (If an answer does not exist, enter DNE.) f(x) = 5 sin(x) + 5 cos(x), 0 ≤ x ≤ 2π (a) Find the interval on which f is increasing. (Enter your answer using interval notation.) Find the interval on which f is decreasing. (Enter your answer using interval notation.) (b) Find the local minimum and maximum values of f. local minimum value local maximum value (c) Find the inflection points. (x, y) = (smaller x-value) (x, y) = (larger x-value) Find the interval on which f is concave up. (Enter your answer using interval notation.) Find the interval on which f is concave down. (Enter your answer using interval notation.)

Answer 1

Answer:

a.  The function is increasing on (-∞, $\frac{\pi}{4}$) and $[\frac{5\pi}{4},$, ∞) and decreasing on $[\frac{\pi}{4},\frac{5\pi}{4}]$

b.  The max point is at $(\frac{\pi}{4},5\sqrt{2})$ and the min point is at $(\frac{5\pi}{4},-5\sqrt{2})$

c.  The inflection points are at $(\frac{3\pi}{4},0),(\frac{7\pi}{4},0)$

d.  The function is concave up on $[\frac{3\pi}{4},\frac{7\pi}{4}]$ and concave down on (-∞, $\frac{3\pi}{4}$] and $[\frac{7\pi}{4},$∞)

Step-by-step explanation:

To find where the function is increasing and decreasing, we need to first find the first derivative of the function, factor it to see where it equals 0, then evaluate the derivative to find the signs of x values before, between, and after those values of 0.  The first derivative is

$f'(x)=5cos(x)-5sin(x)$

The derivative is equal to - then when sin(x) = cos(x).  In the interval between 0 and 2π, these angles are $\frac{\pi}{4},\frac{5\pi}{4}$

You can find these on the unit circle.  You will find at these values, the sin and the cos are identical...same values and same signs.

Now we use these values to determine where our function is increasing and where it is decreasing.  Set up a table:

Interval         -∞<x<$\frac{\pi}{4}$     $\frac{\pi}{4}<x<\frac{5\pi}{4}$ and $\frac{5\pi}{4}<x<$∞

Pick a value first that is between negative infinity and pi/4.  I picked pi/6.  When you plug pi/6 into f'(x) you get a positive value (the value doesn't matter, only the sign does).  This positive value tells us that the function is increasing there from negative infinity up to pi/4.

Pick a value that is between pi/4 and 5pi/4.  I picked pi/2.  When you plug pi/2 into f'(x) you get a negative value.  The function is decreasing from pi/4 to 5pi/4.

Pick a value that is between 5pi/4 and infinity.  I picked 3pi/2.  When you plug 3pi/2 into f'(x) you get a positive value.  The function is increasing over that interval as well.  So the intervals at which the function is increasing and decreasing are,

inc:  (-∞, pi/4], [5pi/4, ∞); dec:  [pi/4, 5pi/4]

The max and min points occur at those values that caused the derivative to go to 0:  pi/4 and 5pi/4.  We evaluate the function at those points to find the x and y values of those max and min points:

$f(\frac{\pi}{4})=5sin(\frac{\pi}{4})+5cos(\frac{\pi}{4})$

Look to the unit circle to find these values:

$f(\frac{\pi}{4})=5(-\frac{\sqrt{2} }{2})+5(\frac{\sqrt{2} }{2})$ and

$f(\frac{\pi}{4})=5\sqrt{2}$ which is a coordinate point of

$(\frac{\pi}{4},5\sqrt{2})$

Next we evaluate 5pi/4 in the function:

$f(\frac{5\pi}{4})=5sin(\frac{5\pi}{4})+5cos(\frac{5\pi}{4})$ so

$f(\frac{5\pi}{4})=-5\sqrt{2}$ and the coordinate point is $(\frac{5\pi}{4},-5\sqrt{2})$

Look at the y-values to see which one is greater, and that is the point that is your max; the other is the min.

For the next part, we find the inflection intervals and their corresponding x/y values in the second derivative, which is

$f''(x)=-5sin(x)-5cos(x)$.  We will find the points of inflection where this second derivative is equal to 0 by factoring.  First factor out the 5:

$f''(x)=5(-sin(x)-cos(x))$ then set it to equal 0:

$0=5(-sin(x)-cos(x))$

Since 5 does not equal 0, then

$0=-sin(x)-cos(x)$ and

$sin(x)=-cos(x)$  OR  $-sin(x)=cos(x)$

On the interval between 0 and 2pi, these 2 places on the unit circle are at

$x=\frac{3\pi}{4},\frac{7\pi}{4}$

Set up a table again with the following intervals:

(-∞, 3pi/4], [3pi/4, 7pi/4], and [7pi/4, ∞) and pick values of x between each interval.  For the first one I picked pi/2.  The second derivative evaluated at pi/2 gave me a negative sign, so the function is concave down on the interval (-∞, 3pi/4].  Next I picked a value of 3pi/2 and evaluated the second derivative.  It gave me a positive sign, so the function is concave up at the interval [3pi/4, 7pi/4].  Next I picked a value of 11pi/6 and evaluated the second derivative at that value and got a negative sign.  The function is concave down at the interval [7pi/4, ∞).

The points of inflection occur at the values where the second derivative equals 0, so

$f(\frac{3\pi}{4})=0$ and $f(\frac{7\pi}{4})=0$

The inflection points are at

$(\frac{3\pi}{4},0) ,(\frac{7\pi}{4},0)$

Phew!