Chapter 5, Problem 12RE

a.

To determine

To find the intervals on which the function is increasing by using analytical method.

Answer to Problem 12RE

The function $y=\sin 3x+\cos 4x$ is increasing in interval $\left(0,0.172\right)$ , $\left(0.99,1.57\right)$ , $\left(2.14,2.96\right)$ and $\left(3.83,4.71\right)$ .

Explanation of Solution

Given:

The function is $y=\sin 3x+\cos 4x,0\le x\le 2\pi$ .

Calculation:

The function is increasing when $f^{\prime }\left(x\right)>0$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3\cos 3x-4\sin 4x\\ 3\cos 3x-4\sin 4x>0\end{array}$

Below is the graph of $f^{\prime }\left(x\right)$

  

From graph it can be observed that there are total eight critical points that is

  $x=0.172,0.99,1.57,2.14,2.96,3.83,4.71,5.59$

Now ,put $x=0.1$ to check whether function is increasing or decreasing in $\left(0,0.172\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3\cos 3x-4\sin 4x\\ f^{\prime }\left(0.1\right)=3\cos 3\left(0.1\right)-4\sin 4\left(0.1\right)\\ f^{\prime }\left(0.1\right)=1.30>0\end{array}$

Therefore $y=\sin 3x+\cos 4x$ is increasing when $x\in \left(0,0.172\right)$

Now ,put $x=0.5$ to check whether function is increasing or decreasing in $\left(0.172,0.99\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3\cos 3x-4\sin 4x\\ f^{\prime }\left(0.5\right)=3\cos 3\left(0.5\right)-4\sin 4\left(0.5\right)\\ f^{\prime }\left(0.5\right)=-3.42<0\end{array}$

Therefore $y=\sin 3x+\cos 4x$ is decreasing when $x\in \left(0.172,0.99\right)$

Now ,put $x=1$ to check whether function is increasing or decreasing in $\left(0.99,1.57\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3\cos 3x-4\sin 4x\\ f^{\prime }\left(1\right)=3\cos 3\left(1\right)-4\sin 4\left(1\right)\\ f^{\prime }\left(1\right)=0.057>0\end{array}$

Therefore $y=\sin 3x+\cos 4x$ is increasing when $x\in \left(0.99,1.57\right)$

Now ,put $x=2$ to check whether function is increasing or decreasing in $\left(1.57,2.14\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3\cos 3x-4\sin 4x\\ f^{\prime }\left(2\right)=3\cos 3\left(2\right)-4\sin 4\left(2\right)\\ f^{\prime }\left(2\right)=-1.07<0\end{array}$

Therefore $y=\sin 3x+\cos 4x$ is decreasing when $x\in \left(1.57,2.14\right)$

Now ,put $x=2$ to check whether function is increasing or decreasing in $\left(2.14,2.96\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3\cos 3x-4\sin 4x\\ f^{\prime }\left(2.4\right)=3\cos 3\left(2.4\right)-4\sin 4\left(2.4\right)\\ f^{\prime }\left(2.4\right)=2.52>0\end{array}$

Therefore $y=\sin 3x+\cos 4x$ is increasing when $x\in \left(2.14,2.96\right)$

Now ,put $x=3$ to check whether function is increasing or decreasing in $\left(2.96,3.83\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3\cos 3x-4\sin 4x\\ f^{\prime }\left(3\right)=3\cos 3\left(3\right)-4\sin 4\left(3\right)\\ f^{\prime }\left(3\right)=-0.58<0\end{array}$

Therefore $y=\sin 3x+\cos 4x$ is decreasing when $x\in \left(2.96,3.83\right)$

Now ,put $x=4$ to check whether function is increasing or decreasing in $\left(3.83,4.71\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3\cos 3x-4\sin 4x\\ f^{\prime }\left(4\right)=3\cos 3\left(4\right)-4\sin 4\left(4\right)\\ f^{\prime }\left(4\right)=3.68>0\end{array}$

Therefore $y=\sin 3x+\cos 4x$ is increasing when $x\in \left(3.83,4.71\right)$

Now ,put $x=5$ to check whether function is increasing or decreasing in $\left(4.71,5.59\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3\cos 3x-4\sin 4x\\ f^{\prime }\left(5\right)=3\cos 3\left(5\right)-4\sin 4\left(5\right)\\ f^{\prime }\left(5\right)=-5.9<0\end{array}$

Therefore $y=\sin 3x+\cos 4x$ is decreasing when $x\in \left(4.71,5.59\right)$

hence, the function $y=\sin 3x+\cos 4x$ is increasing in interval $\left(0,0.172\right)$ , $\left(0.99,1.57\right)$ , $\left(2.14,2.96\right)$ and $\left(3.83,4.71\right)$ .

b.

To determine

To find the intervals on which the function is decreasing by using analytical method.

Answer to Problem 12RE

The function $y=\sin 3x+\cos 4x$ is decreasing in interval $\left(0.172,0.99\right)$ , $\left(1.57,2.14\right)$ , $\left(2.96,3.83\right)$ and $\left(4.71,5.59\right)$ .

Explanation of Solution

Given:

The function is $y=\sin 3x+\cos 4x,0\le x\le 2\pi$ .

Calculation:

The function is decreasing when $f^{\prime }\left(x\right)<0$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3\cos 3x-4\sin 4x\\ 3\cos 3x-4\sin 4x>0\end{array}$

Below is the graph of $f^{\prime }\left(x\right)$

  

From graph it can be observed that there are total eight critical points that is

  $x=0.172,0.99,1.57,2.14,2.96,3.83,4.71,5.59$

Now ,put $x=0.1$ to check whether function is increasing or decreasing in $\left(0,0.172\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3\cos 3x-4\sin 4x\\ f^{\prime }\left(0.1\right)=3\cos 3\left(0.1\right)-4\sin 4\left(0.1\right)\\ f^{\prime }\left(0.1\right)=1.30>0\end{array}$

Therefore $y=\sin 3x+\cos 4x$ is increasing when $x\in \left(0,0.172\right)$

Now ,put $x=0.5$ to check whether function is increasing or decreasing in $\left(0.172,0.99\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3\cos 3x-4\sin 4x\\ f^{\prime }\left(0.5\right)=3\cos 3\left(0.5\right)-4\sin 4\left(0.5\right)\\ f^{\prime }\left(0.5\right)=-3.42<0\end{array}$

Therefore $y=\sin 3x+\cos 4x$ is decreasing when $x\in \left(0.172,0.99\right)$

Now ,put $x=1$ to check whether function is increasing or decreasing in $\left(0.99,1.57\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3\cos 3x-4\sin 4x\\ f^{\prime }\left(1\right)=3\cos 3\left(1\right)-4\sin 4\left(1\right)\\ f^{\prime }\left(1\right)=0.057>0\end{array}$

Therefore $y=\sin 3x+\cos 4x$ is increasing when $x\in \left(0.99,1.57\right)$

Now ,put $x=2$ to check whether function is increasing or decreasing in $\left(1.57,2.14\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3\cos 3x-4\sin 4x\\ f^{\prime }\left(2\right)=3\cos 3\left(2\right)-4\sin 4\left(2\right)\\ f^{\prime }\left(2\right)=-1.07<0\end{array}$

Therefore $y=\sin 3x+\cos 4x$ is decreasing when $x\in \left(1.57,2.14\right)$

Now ,put $x=2$ to check whether function is increasing or decreasing in $\left(2.14,2.96\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3\cos 3x-4\sin 4x\\ f^{\prime }\left(2.4\right)=3\cos 3\left(2.4\right)-4\sin 4\left(2.4\right)\\ f^{\prime }\left(2.4\right)=2.52>0\end{array}$

Therefore $y=\sin 3x+\cos 4x$ is increasing when $x\in \left(2.14,2.96\right)$

Now ,put $x=3$ to check whether function is increasing or decreasing in $\left(2.96,3.83\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3\cos 3x-4\sin 4x\\ f^{\prime }\left(3\right)=3\cos 3\left(3\right)-4\sin 4\left(3\right)\\ f^{\prime }\left(3\right)=-0.58<0\end{array}$

Therefore $y=\sin 3x+\cos 4x$ is decreasing when $x\in \left(2.96,3.83\right)$

Now ,put $x=4$ to check whether function is increasing or decreasing in $\left(3.83,4.71\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3\cos 3x-4\sin 4x\\ f^{\prime }\left(4\right)=3\cos 3\left(4\right)-4\sin 4\left(4\right)\\ f^{\prime }\left(4\right)=3.68>0\end{array}$

Therefore $y=\sin 3x+\cos 4x$ is increasing when $x\in \left(3.83,4.71\right)$

Now ,put $x=5$ to check whether function is increasing or decreasing in $\left(4.71,5.59\right)$

  $\begin{array}{l}{f}^{\prime }\left(x\right)=3\cos 3x-4\sin 4x\\ f^{\prime }\left(5\right)=3\cos 3\left(5\right)-4\sin 4\left(5\right)\\ f^{\prime }\left(5\right)=-5.9<0\end{array}$

Therefore $y=\sin 3x+\cos 4x$ is decreasing when $x\in \left(4.71,5.59\right)$

hence, the function $y=\sin 3x+\cos 4x$ is decreasing in interval $\left(0.172,0.99\right)$ , $\left(1.57,2.14\right)$ , $\left(2.96,3.83\right)$ and $\left(4.71,5.59\right)$ .

c.

To determine

To find the intervals on which the function is concave up by using analytical method.

Answer to Problem 12RE

The function $y=\sin 3x+\cos 4x$ is concave up in the interval $\left(0.54,1.26\right)$ , $\left(1.87,2.59\right)$ , $\left(3.4,4.28\right)$ and $\left(5.14,5.99\right)$ .

Explanation of Solution

Given:

The function is $y=\sin 3x+\cos 4x,0\le x\le 2\pi$ .

Calculation:

The graph of a twice differentiable function $y=f\left(x\right)$ is

Concave up on any interval where $f^{″}\left(x\right)>0$ and concave down on any interval where $f^{″}\left(x\right)<0$

Since, $y=\sin 3x+\cos 4x$

First derivative : $y^{\prime }=3\cos 3x-4\sin 4x$

Second derivative : $y^{″}=-9\sin 3x-16\cos 4x$

below is the graph of $y^{″}=-9\sin 3x-16\cos 4x$

  

From graph it is clear that , there are total eight critical points that are

  $x=0.54,1.26,1.87,2.59,3.4,4.28,5.14,5.99$

Now, put $x=0.1$ to check whether function is concave up or down in the interval $\left(0,0.54\right)$

  $\begin{array}{l}{f}^{″}(x)=-9\sin 3x-16\cos 4x\\ f^{″}(0.2)=-9\sin 3\left(0.2\right)-16\cos 4\left(0.2\right)\\ f^{″}(0.2)=-16.22<0\end{array}$

Therefore, $y=\sin 3x+\cos 4x$ is concave down in interval $\left(0,0.54\right)$

Now, put $x=1$ to check whether function is concave up or down in the interval $\left(0.54,1.26\right)$

  $\begin{array}{l}{f}^{″}(x)=-9\sin 3x-16\cos 4x\\ f^{″}(1)=-9\sin 3\left(1\right)-16\cos 4\left(1\right)\\ f^{″}(1)=9.18>0\end{array}$

Therefore, $y=\sin 3x+\cos 4x$ is concave up in interval $\left(0.54,1.26\right)$

Now, put $x=1.5$ to check whether function is concave up or down in the interval $\left(1.26,1.87\right)$

  $\begin{array}{l}{f}^{″}(x)=-9\sin 3x-16\cos 4x\\ f^{″}(1.5)=-9\sin 3\left(1.5\right)-16\cos 4\left(1.5\right)\\ f^{″}(1.5)=-6.5<0\end{array}$

Therefore, $y=\sin 3x+\cos 4x$ is concave down in interval $\left(1.26,1.87\right)$

Now, put $x=2$ to check whether function is concave up or down in the interval $\left(1.87,2.59\right)$

  $\begin{array}{l}{f}^{″}(x)=-9\sin 3x-16\cos 4x\\ f^{″}(2)=-9\sin 3\left(2\right)-16\cos 4\left(2\right)\\ f^{″}(2)=4.8>0\end{array}$

Therefore, $y=\sin 3x+\cos 4x$ is concave up in interval $\left(1.87,2.59\right)$

Now, put $x=3$ to check whether function is concave up or down in the interval $\left(2.59,3.4\right)$

  $\begin{array}{l}{f}^{″}(x)=-9\sin 3x-16\cos 4x\\ f^{″}(3)=-9\sin 3\left(3\right)-16\cos 4\left(3\right)\\ f^{″}(3)=-17.2<0\end{array}$

Therefore, $y=\sin 3x+\cos 4x$ is concave down in interval $\left(2.59,3.4\right)$

Now, put $x=4$ to check whether function is concave up or down in the interval $\left(3.4,4.28\right)$

  $\begin{array}{l}{f}^{″}(x)=-9\sin 3x-16\cos 4x\\ f^{″}(4)=-9\sin 3\left(4\right)-16\cos 4\left(4\right)\\ f^{″}(4)=20.15>0\end{array}$

Therefore, $y=\sin 3x+\cos 4x$ is concave up in interval $\left(3.4,4.28\right)$

Now, put $x=5$ to check whether function is concave up or down in the interval $\left(4.28,5.14\right)$

  $\begin{array}{l}{f}^{″}(x)=-9\sin 3x-16\cos 4x\\ f^{″}(5)=-9\sin 3\left(5\right)-16\cos 4\left(5\right)\\ f^{″}(5)=-12.38<0\end{array}$

Therefore, $y=\sin 3x+\cos 4x$ is concave down in interval $\left(4.28,5.14\right)$

Now, put $x=5.2$ to check whether function is concave up or down in the interval $\left(5.14,5.99\right)$

  $\begin{array}{l}{f}^{″}(x)=-9\sin 3x-16\cos 4x\\ f^{″}(5.2)=-9\sin 3\left(5.2\right)-16\cos 4\left(5.2\right)\\ f^{″}(5.2)=4.95>0\end{array}$

Therefore, $y=\sin 3x+\cos 4x$ is concave up in interval $\left(5.14,5.99\right)$

Now, put $x=6$ to check whether function is concave up or down in the interval $\left(5.99,2\pi \right)$

  $\begin{array}{l}{f}^{″}(x)=-9\sin 3x-16\cos 4x\\ f^{″}(6)=-9\sin 3\left(6\right)-16\cos 4\left(6\right)\\ f^{″}(6)=-0.02<0\end{array}$

Therefore, $y=\sin 3x+\cos 4x$ is concave down in interval $\left(5.99,2\pi \right)$

Hence, the function $y=\sin 3x+\cos 4x$ is concave up in the interval $\left(0.54,1.26\right)$ , $\left(1.87,2.59\right)$ , $\left(3.4,4.28\right)$ and $\left(5.14,5.99\right)$ .

d.

To determine

To find the intervals on which the function is concave down by using analytical method.

Answer to Problem 12RE

The function $y=\sin 3x+\cos 4x$ is concave down in the interval $\left(0,0.54\right)$ , $\left(1.26,1.87\right)$ , $\left(2.59,3.4\right)$ and $\left(4.28,5.14\right)$

Explanation of Solution

Given:

The function is $y=\sin 3x+\cos 4x,0\le x\le 2\pi$ .

Calculation:

The graph of a twice differentiable function $y=f\left(x\right)$ is

Concave up on any interval where $f^{″}\left(x\right)>0$ and concave down on any interval where $f^{″}\left(x\right)<0$

Since, $y=\sin 3x+\cos 4x$

First derivative : $y^{\prime }=3\cos 3x-4\sin 4x$

Second derivative : $y^{″}=-9\sin 3x-16\cos 4x$

below is the graph of $y^{″}=-9\sin 3x-16\cos 4x$

  

From graph it is clear that , there are total eight critical points that are

  $x=0.54,1.26,1.87,2.59,3.4,4.28,5.14,5.99$

Now, put $x=0.1$ to check whether function is concave up or down in the interval $\left(0,0.54\right)$

  $\begin{array}{l}{f}^{″}(x)=-9\sin 3x-16\cos 4x\\ f^{″}(0.2)=-9\sin 3\left(0.2\right)-16\cos 4\left(0.2\right)\\ f^{″}(0.2)=-16.22<0\end{array}$

Therefore, $y=\sin 3x+\cos 4x$ is concave down in interval $\left(0,0.54\right)$

Now, put $x=1$ to check whether function is concave up or down in the interval $\left(0.54,1.26\right)$

  $\begin{array}{l}{f}^{″}(x)=-9\sin 3x-16\cos 4x\\ f^{″}(1)=-9\sin 3\left(1\right)-16\cos 4\left(1\right)\\ f^{″}(1)=9.18>0\end{array}$

Therefore, $y=\sin 3x+\cos 4x$ is concave up in interval $\left(0.54,1.26\right)$

Now, put $x=1.5$ to check whether function is concave up or down in the interval $\left(1.26,1.87\right)$

  $\begin{array}{l}{f}^{″}(x)=-9\sin 3x-16\cos 4x\\ f^{″}(1.5)=-9\sin 3\left(1.5\right)-16\cos 4\left(1.5\right)\\ f^{″}(1.5)=-6.5<0\end{array}$

Therefore, $y=\sin 3x+\cos 4x$ is concave down in interval $\left(1.26,1.87\right)$

Now, put $x=2$ to check whether function is concave up or down in the interval $\left(1.87,2.59\right)$

  $\begin{array}{l}{f}^{″}(x)=-9\sin 3x-16\cos 4x\\ f^{″}(2)=-9\sin 3\left(2\right)-16\cos 4\left(2\right)\\ f^{″}(2)=4.8>0\end{array}$

Therefore, $y=\sin 3x+\cos 4x$ is concave up in interval $\left(1.87,2.59\right)$

Now, put $x=3$ to check whether function is concave up or down in the interval $\left(2.59,3.4\right)$

  $\begin{array}{l}{f}^{″}(x)=-9\sin 3x-16\cos 4x\\ f^{″}(3)=-9\sin 3\left(3\right)-16\cos 4\left(3\right)\\ f^{″}(3)=-17.2<0\end{array}$

Therefore, $y=\sin 3x+\cos 4x$ is concave down in interval $\left(2.59,3.4\right)$

Now, put $x=4$ to check whether function is concave up or down in the interval $\left(3.4,4.28\right)$

  $\begin{array}{l}{f}^{″}(x)=-9\sin 3x-16\cos 4x\\ f^{″}(4)=-9\sin 3\left(4\right)-16\cos 4\left(4\right)\\ f^{″}(4)=20.15>0\end{array}$

Therefore, $y=\sin 3x+\cos 4x$ is concave up in interval $\left(3.4,4.28\right)$

Now, put $x=5$ to check whether function is concave up or down in the interval $\left(4.28,5.14\right)$

  $\begin{array}{l}{f}^{″}(x)=-9\sin 3x-16\cos 4x\\ f^{″}(5)=-9\sin 3\left(5\right)-16\cos 4\left(5\right)\\ f^{″}(5)=-12.38<0\end{array}$

Therefore, $y=\sin 3x+\cos 4x$ is concave down in interval $\left(4.28,5.14\right)$

Now, put $x=5.2$ to check whether function is concave up or down in the interval $\left(5.14,5.99\right)$

  $\begin{array}{l}{f}^{″}(x)=-9\sin 3x-16\cos 4x\\ f^{″}(5.2)=-9\sin 3\left(5.2\right)-16\cos 4\left(5.2\right)\\ f^{″}(5.2)=4.95>0\end{array}$

Therefore, $y=\sin 3x+\cos 4x$ is concave up in interval $\left(5.14,5.99\right)$

Now, put $x=6$ to check whether function is concave up or down in the interval $\left(5.99,2\pi \right)$

  $\begin{array}{l}{f}^{″}(x)=-9\sin 3x-16\cos 4x\\ f^{″}(6)=-9\sin 3\left(6\right)-16\cos 4\left(6\right)\\ f^{″}(6)=-0.02<0\end{array}$

Therefore, $y=\sin 3x+\cos 4x$ is concave down in interval $\left(5.99,2\pi \right)$

Hence, the function $y=\sin 3x+\cos 4x$ is concave down in the interval $\left(0,0.54\right)$ , $\left(1.26,1.87\right)$ , $\left(2.59,3.4\right)$ , $\left(4.28,5.14\right)$ and $\left(5.99,2\pi \right)$ .

e.

To determine

To find any local extreme values.

Answer to Problem 12RE

  $y=\sin 3x+\cos 4x$ has local maximum value at point $\left(0.176,1.266\right),\left(1.57,0\right),\left(2.965,1.266\right),\left(4.71,2\right)\text{ and }\left(6.28,1\right)$ and local minimum value at point $\left(0,1\right),\left(0.99,-0.51\right),\left(3.83,-1.80\right)\text{ and }\left(5.59,-1.80\right)$ .

Explanation of Solution

Given:

The function is $y=\sin 3x+\cos 4x,0\le x\le 2\pi$ .

Calculation:

Below is the graph of $y=\sin 3x+\cos 4x$

  

From graph it is clear that $y=\sin 3x+\cos 4x$ has local maximum value at point $\left(0.176,1.266\right),\left(1.57,0\right),\left(2.965,1.266\right),\left(4.71,2\right)\text{ and }\left(6.28,1\right)$ and local minimum value at point $\left(0,1\right),\left(0.99,-0.51\right),\left(3.83,-1.80\right)\text{ and }\left(5.59,-1.80\right)$

f.

To determine

To find inflections points.

Answer to Problem 12RE

The inflection points are $x=0.54,1.26,1.87,2.59,3.4,4.28,5.14,5.99$

Explanation of Solution

Given:

The function is $y=\sin 3x+\cos 4x,0\le x\le 2\pi$ .

Calculation:

Inflection point of any function is a point where the graph of function has a tangent line and where the concavity changes.

Since, the intervals in which function is concave up are $\left(0.54,1.26\right)$ , $\left(1.87,2.59\right)$ , $\left(3.4,4.28\right)$ and $\left(5.14,5.99\right)$ .

The intervals in which function is concave down are $\left(0,0.54\right)$ , $\left(1.26,1.87\right)$ , $\left(2.59,3.4\right)$ , $\left(4.28,5.14\right)$ and $\left(5.99,2\pi \right)$

Therefore, the inflection points are $x=0.54,1.26,1.87,2.59,3.4,4.28,5.14,5.99$ .