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EleoNora [17]
3 years ago
13

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

Mathematics
2 answers:
vlabodo [156]3 years ago
5 0

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.

Ad libitum [116K]3 years ago
3 0
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, ...
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