Question

For the function y=-1+6 cos(2pi/7(x-5)) what is the minimum value?

Answer 1

Answer:

The value of y lie in the interval [-7,5] and the minimum value of given function is -7.

Step-by-step explanation:

The given function is

$y=-1+6\cos (\frac{2\pi}{7}(x-5))$

We know that the value of cosine function lies between -1 and 1.

By the above property of cosine function we get

$-1\leq \cos (\frac{2\pi}{7}(x-5))\leq 1$

Multiply 6 each side.

$-6\leq 6\cos (\frac{2\pi}{7}(x-5))\leq 6$

Subtract 1 from each side.

$-6-1\leq 6\cos (\frac{2\pi}{7}(x-5))-1\leq 6-1$

$-7\leq y\leq 5$                             $[\because y=-1+6\cos (\frac{2\pi}{7}(x-5))]$

Therefore the value of y lie in the interval [-7,5] and the minimum value of given function is -7.

Answer 2

Usually one will differentiate the function to find the minimum/maximum point, but in this case differentiating yields:
$\frac{2\pi}{7(x-5)^{2}}\sin{\frac{2\pi}{7(x-5)}}}$
which contains multiple solution if one tries to solve for x when the differentiated form is 0.

I would, though, venture a guess that the minimum value would be (approaching) 5, since the function would be undefined in the vicinity.

If, however, the function is
$-1+\cos{\frac{2\pi}{7}(x-5)}}$
Then differentiating and equating to 0 yields:
$\sin{\frac{2\pi}{7}(x-5)}}=0$
which gives:
$x=5$ or $8.5$

We reject x=5 as it is when it ix the maximum and thus,
$x=8.5\pm7n$, for $n=0,\pm 1,\pm 2, ...$