Question

a) Find the critical points, if any, of the following function on the given interval.

b) Determine the absolute extreme values of f on the given interval.

c) Use a graphing utility to confirm your conclusions.

f(x)=sin(3x)on [−π4,π3]

I Know f′(x)=3cos(3x).

However, I'm not exactly sure how to determine the zeros for x for a trig function.

and even I know how to find the zeros of a cosine function I don't now what to do with them after that.

please if somebody can solve this I've been at it for over 5 hr's

Answer 1

Question #36767, Calculus

a) Find the critical points, if any, of the following function on the given interval.

b) Determine the absolute extreme values of $f$ on the given interval.

c) Use a graphing utility to confirm your conclusions.

$$f(x) = \sin(3x) \quad \text{on } [\neg 4, 3]$$

I Know $f'(x) = 3\cos(3x)$.

However, I'm not exactly sure how to determine the zeros for $x$ for a trig function.

and even I know how to find the zeros of a cosine function I don't know what to do with them after that.

please if somebody can solve this I've been at it for over 5 hr's

Solution

1. $f(x) = \sin(3x)$ on $[\neg 4, 3]$

a) To determine the critical value(s) we must first find $f'(x)$:

$$f'(x) = 3\cos(3x).$$

Next we must find value(s) of $c$ such that $f'(c) = 0$.

$$\text{So, } f'(x) = 0 \iff 3\cos(3x) = 0 \iff \cos(3x) = 0.$$

The last equation has the general solution $3x = \cos^{-1}(0) + \pi n = \pi/2 + \pi n$, where $n$ is an integer (including zero). Hence, $x = \pi/6 + \pi n/3$, where $n$ is an integer.

For $\neg 4 \leq x \leq 3$ we have $-12 \leq n \leq 8$, i.e., 21 critical points for given interval.

b) By setting $n = 2k$ we get $x = \pi/6 + 2\pi k/3$, for $-6 \leq k \leq 4$ or $x = -23\pi/6, -19\pi/6, -5\pi/2, -11\pi/6, -7\pi/6, -\pi/2, \pi/6, 5\pi/6, 3\pi/2, 13\pi/6, 17\pi/6$.

And then we find the function values at these critical points

$$f(\pi/6 + 2\pi k/3) = \sin(\pi/2 + 2\pi k) = 1.$$

By setting $n = 2k - 1$ we obtain $x = -\pi/6 + 2\pi k/3$, for $-5 \leq k \leq 4$ or $x = -21\pi/6, -17\pi/6, -13\pi/6, -3\pi/2, -5\pi/6, -\pi/6, \pi/2, 7\pi/6, 11\pi/6, 15\pi/6$.

For these critical points we get

$$f(-\pi/6 + 2\pi k/3) = \sin(-\pi/2 + 2\pi k) = -1.$$

The values of the function at the end points of interval are

$$f(-\pi 4) = \sin(-12\pi) = 0, \quad f(\pi 3) = \sin(9\pi) = 0.$$

Answers

a) The critical points are $x = -\pi/6 + 2\pi k/3$, for $-5 \leq k \leq 4$ and $x = \pi/6 + 2\pi k/3$, for $-6 \leq k \leq 4$.

b) The absolute minimum occurs at the critical values $x = -\pi / 6 + 2\pi k / 3$ , for $-5 \leq k \leq 4$ and is equal to $-1$ ; the absolute maximum occurs at the critical values $x = \pi / 6 + 2\pi k / 3$ , for $-6 \leq k \leq 4$ and is equal to 1.

c) We use SpeQ 3.4 graphing utility to confirm our conclusions.

2. $f(x) = \sin (3x)$ on $[- \pi / 4, \pi / 3]$

a) To determine the critical value(s) we must first find $f'(x)$ :

$f^{\prime}(x) = 3\cos (3x)$

Next we must find value(s) of c such that $f'(c) = 0$ .

So, $f'(x) = 0 \Leftrightarrow 3\cos(3x) = 0 \Leftrightarrow \cos(3x) = 0$ .

The last equation has the general solution $3x = \cos^{-1}(0) + \pi n = \pi / 2 + \pi n$ , where $n$ is an integer (including zero). Hence, $x = \pi / 6 + \pi n / 3$ , where $n$ is an integer.

For $-\pi /4\leq x\leq \pi /3$ we have $n = -1,0,$ i.e., two critical points for the given interval $x = \pm \pi /6$

b) First, we find the function values at these critical points

$f(\pi /6) = \sin (\pi /2) = 1,$

$f(-\pi /6) = \sin (-\pi /2) = -1.$

The values of the function at the end points of interval are

$f(-\pi /4) = \sin (-3\pi /4) = -\frac{\sqrt{2}}{2},f(\pi /3) = \sin (\pi) = 0.$

Answers

a) The critical points are $x = -\pi / 6$ and $x = \pi / 6$ .

b) The absolute minimum occurs at the critical value $x = -\pi / 6$ and is equal to $-1$ ; the absolute maximum occurs at the critical value $\pi / 6$ and is equal to 1.

c) We use SpeQ 3.4 graphing utility to confirm our conclusions.