Maximize $\cos(2x) - e^{3x}$

Question

I'm not certain how to maximize this function because I don't know how to solve $\frac{df}{dx}=0$. However, as $x \rightarrow -\infty$ then $e^{3x} \rightarrow 0$ and so $\cos(x)$ can achieve maximal value of 1. I'm not sure how to express this (assuming it's correct). Below is a graph of $\cos(x) - e^x$ which illustrates that as $x$ increases, $e^x$ increases monotonically and "pulls" the maximum value of $f(x)$ down.

$\cos(x) - \exp(x)$">

Answer 1

Take the first derivative of $f(x) = \cos(2x) - e^{3x}$, $f'(x) = -2\sin(2x)-3e^{3x}$. Note that, $f(0)=0$ and $f'(x)<0$ for $x>0$. So $f(x)<0$ for all $x>0$. Rest of it is as you observed. However, this function does not have a maximum. It has a supremum which turns out to be $1$.

Answer 2

The function does not have a unique maximum. As $x\rightarrow -\infty $, the function oscillates on the interval $\left [ -,1 \right ]$. As $x\rightarrow \infty $, the function approaches $-\infty $.

$$f\left ( x \right )=\cos\left ( 2x \right )-e^{3x}$$

$$f^{'}\left ( x \right )=-2\sin\left ( 2x \right )-3e^{3x}$$

$f^{'}\left ( x \right )=0$ if $-2\sin\left ( 2x \right )=3e^{3x}$ Note that $f\left ( x \right )$ has infinitely many critical points. Except the first one, the rest occurs periodically. Please take a look at the graph below. $-2\sin\left ( 2x \right )$ and $3e^{3x}$ intersect each other at infinitely many points on $\left (-\infty ,0 \right ]$.

Due to the effect of $-e^{3x}$, the critical points do not exactly happen periodically either, at least not near $x=0$.

However, as $x\rightarrow -\infty $, $-e^{3x}$ loses its effect and the entire function becomes periodic again. As the table shows, as $\mathbf{x\rightarrow -\infty }$ , the function $f\left ( x \right )=\cos\left ( 2x \right )-e^{3x}$ reaches its upper bound at $1$ at $-n \pi$, for $n=1,2,3,...$.

Answer 3

We can have quite good approximations of the values of the maxima as well as the locations where they do happen.

As @user709318 did show, the maximum occur around $x=-k \pi$. So, develop as a Taylor series around $x=-k \pi$ to get

$$y=\cos(2x) - e^{3x}=\left(1-e^{-3 \pi k}\right)- 3 e^{-3 \pi k}(x+k\pi)- \left(2+\frac{9}{2} e^{-3 \pi k} \right)(x+k\pi )^2+O\left((x+k\pi)^3\right)$$ Differentiating and solving for $x$ gives the maximum to be located at $$x_k=-k\pi-\frac{3}{9+4 e^{3 \pi k}}$$ for which the maximum is given by $$y_k=1-\frac{8+9 e^{-3 \pi k}}{8 e^{3 \pi k}+18}$$

Computed with hight accuracy, some results

$$\left( \begin{array}{cccc} & y & x & y & x \\ k & \text{approximation}& \text{approximation}& \text{exact}& \text{exact}\\ 0 & 0.34615384615385 & -0.23076923076923 & 0.40692125228158& -0.25384615384615 \\ -1 & 0.99991930780756 & -3.14165316724027 & 0.99991930780756& -3.14165317481513 \\ -2 & 0.99999999348759 & -6.28318531206390 & 0.99999999348759& -6.28318531770143 \\ -3 & 0.99999999999947 & -9.42477796076977 & 0.99999999999947& -9.42477796076977 \\ -4 & 1.00000000000000 & -12.5663706143592 & 0.99999999999999& -12.5663706870126 \\ -5 & 1.00000000000000 & -15.7079632679490 & 0.99999999999998& -15.7079633637301 \\ -6 & 1.00000000000000 & -18.8495559215388 & 0.99999999999997& -18.8495560467538 \end{array} \right)$$