# Whats the maximum value of $y=6\cos\left( \frac {2\pi}{14} x\right)-2?$

Please show the correct way how to do this thanks. I got this..

$y$ will be maximum when $\cos 2\pi/14x - 2$ is maximum i. e. when $2\pi/14x- 2 = 0$ [ $\cos 0$ is maximum $= 1$ ] so or $$2\pi/14x = 2$$ or $$14x = \pi$$ or $$x= \pi/14$$ so maximum value of y = 6 ans ---------------- alternatively $$\frac{dy}{dx} = 6$$ ( $-\sin(2\pi/14x -2)) *2pi/14( -1/x^2)$ for maxima & minima $$\frac{dy}{dx} = 0$$ so $$6( - \sin 2\pi/14x - 2 )* 2pi/14* (-1/x^2) = 0$$ $$\sin (2\pi/14x) - 2 =0$$ $$2\pi/14x = 2$$ $$x = \pi/14$$

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Well the maximum of $6\cos\frac{2\pi}{14}x$ should just be the maximum of $6\cos x$, i.e. $6$. – oldrinb May 6 '13 at 1:54
The maximum point is when the gradiet dy/dx = 0. Therefore, differentiate y = 6 cos (2pi/14)x - 2 and subsititute dy/dx = 0 to obtain the value x. You look at a standard cosine waveform and you will see that the peak (maximum) or the negative peak (minimum) will always be the point when gradient = 0 (i.e the point of change from positive to negative gradient and vice versa) – user73122 May 6 '13 at 1:55
The cosine of $t$ is $1$ when $t$ is a multiple of $2\pi$, and never bigger. So our function reaches a max of $6-2$ at $x=14n$, where $n$ is any integer. – André Nicolas May 6 '13 at 2:12
@user73122: Is the vertical bar some part of the problem with an absolute value or just a typo? – Amzoti May 6 '13 at 2:14
Please use enough parentheses to be unambiguous. When you write 2pi/14x you probably mean (2pi/14)x but other times people write the same form meaning 2pi/(14x). Even better, use $\LaTeX$, and stack your fractions. To get started, you could look here – Ross Millikan May 6 '13 at 3:31

Assume the expression is:

$$y = 6cos(\frac{2\pi}{14}x) - 2$$

I showed you how to solve the problem in two ways - both "trigonometry" and "calculus" methods. This should help you compare the results from both of the methods.

"Trigonometry" method

Use this form:

$y = \alpha cos(\beta x) + \delta$ where:

• $\alpha$ is the amplitude
• $\beta$ is the variable for the period of the function, which is $|\frac{2\pi}{\beta}|$.
• $\delta$ is the vertical phase shift.

Suppose we focus on $y = 6cos(\frac{2\pi}{14}x)$. We know that the amplitude of that function is 6. This means that the graph of the function constantly oscillates within y = 6 and y = -6. With oldrinb's comment, The highest point the graph of the function can reach has the y-coordinate 6.

Now, if we subtract that function by 2, then the maximum value decreases by 2. For this case, the maximum value is 4.

"Calculus" method

Recall that:

$d/dx [\text{cos}(f(x))] = -f'(x)\text{sin}(f(x))$ [$\gamma$]

Let $f(x) = \frac{2\pi x}{14}$ which is equivalent to $\frac{\pi x}{7}$. Then, $f'(x) = \frac{\pi}{7}$ So using Equation [$\gamma$], we obtain:

$$dy/dx = -6 \cdot \frac{\pi}{7}\text{sin}(\frac{\pi x}{7}) - 0$$ $$= \frac{-6\pi}{7}\text{sin}(\frac{\pi x}{7})$$

Let $\frac{dy}{dx} = 0$. Then:

$$0 = \frac{-6\pi}{7}\text{sin}(\frac{\pi x}{7})$$ $$0 = \text{sin}(\frac{\pi x}{7})$$ $$arcsin(0) = \frac{\pi x}{7}$$ $$\pi k = \frac{\pi x}{7}, k \in \mathbb{Z}$$ So $x = 7k$ for $k \in \mathbb{Z}$.

Rewrite the expression as follows:

$$x = 14l + 7, l \in \mathbb{Z}$$ $$x = 14m, m \in \mathbb{Z}$$

Use either the first or second derivative test to determine whether each point is a min or max (I leave that up to you to work this out.). You should get the max at $x = 14l + 7$ where $l \in \mathbb{Z}$. Finally, substitute that value for the original function. Thus, you get the maximum value of 4.

Otherwise, if the expression is $y = 6\text{cos}(\frac{2\pi x}{14} - 2)$, then the maximum value is actually 6 as what oldrinb said. The expression inside cosine doesn't affect the maximum and minimum values of the expression. The only variables that affect these values are the amplitudes and the vertical phase shift.

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$6\cos\frac{2\pi x}{14}-2$ will be maximum when $\cos\frac{2\pi x}{14}$ attains maximum value.

Now, for real value of $\theta, -1\le \cos\theta\le 1$

So, the maximum value of $\cos\theta$ is $1$ which occurs if $\theta=2n\pi$ where $n$ is any integer

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I.e., when $2\pi x/14 = 2n\pi$ or $x = 14n$. – marty cohen May 6 '13 at 5:37
@martycohen, yes – lab bhattacharjee May 6 '13 at 5:55
I love short answers with no-nonsense. +1 – Squirtle Dec 23 '13 at 23:02