1. ## [SOLVED] Trignometry Help!

Question:
Find the maximum value and minimum value of each of the following functions. In each case, give the least positive value of x at which they occur.

i. $\displaystyle 9 + \sin(4x-20)^o$

ii. $\displaystyle \frac{30}{11 - 5\cos(\frac{1}{2}x-45)^o}$

Attempt:

i. Maximum Value $\displaystyle = 9+1 = 10$ , Minimum Value $\displaystyle = 9-1=8$

ii. Maximum Value = $\displaystyle \frac{30}{11-5\times1} = 5$ , Minimum Value = $\displaystyle \frac{30}{11-5\times-1} = 1.875$

How can I get the values of x in both questions?

2. Originally Posted by looi76
Question:
Find the maximum value and minimum value of each of the following functions. In each case, give the least positive value of x at which they occur.

i. $\displaystyle 9 + \sin(4x-20)^o$

ii. $\displaystyle \frac{30}{11 - 5\cos(\frac{1}{2}x-45)^o}$

Attempt:

i. Maximum Value $\displaystyle = 9+1 = 10$ , Minimum Value $\displaystyle = 9-1=8$

ii. Maximum Value = $\displaystyle \frac{30}{11-5\times1} = 5$ , Minimum Value = $\displaystyle \frac{30}{11-5\times-1} = 1.875$

How can I get the values of x in both questions?
You just need to solve the trig sections for +/-1

ie
$\displaystyle 1 = \sin(4x-20)$

$\displaystyle -1 = \sin(4x-20)$

$\displaystyle 1= \cos(\frac{1}{2}x-45)$

$\displaystyle -1= \cos(\frac{1}{2}x-45)$

Now there are multiple solutions to these, but choose the smallest positive one. It should be clear that when you solve, for example, the first one, that sin(4x-20) will then be equal to one, it's largest possible y-value. And this is corresponds to the "1" you substituted into your equation in the solution you have already provided (9 + 1 = 10)

3. i. Maximum Value

$\displaystyle Sin(4x-20) = 1$
$\displaystyle 4x - 20 = Sin^{-1}(1)$
$\displaystyle 4x - 20 = 90$
$\displaystyle 4x = 90 + 20$
$\displaystyle 4x = 110$
$\displaystyle x = \frac{110}{4}$
$\displaystyle x = 27.5$

Minimum Value

$\displaystyle Sin(4x-20) = -1$
$\displaystyle 4x - 20 = Sin^{-1}(-1)$
$\displaystyle 4x - 20 = -90$
$\displaystyle 4x = -90 + 20$
$\displaystyle 4x = -70$
$\displaystyle x = \frac{-70}{4}$
$\displaystyle x = -17.5$

Why is the minimum value is wrong?

4. Originally Posted by looi76
i. Maximum Value

$\displaystyle Sin(4x-20) = 1$
$\displaystyle 4x - 20 = Sin^{-1}(1)$
$\displaystyle 4x - 20 = 90$
$\displaystyle 4x = 90 + 20$
$\displaystyle 4x = 110$
$\displaystyle x = \frac{110}{4}$
$\displaystyle x = 27.5$

Minimum Value

$\displaystyle Sin(4x-20) = -1$
$\displaystyle 4x - 20 = Sin^{-1}(-1)$
$\displaystyle 4x - 20 = -90$
$\displaystyle 4x = -90 + 20$
$\displaystyle 4x = -70$
$\displaystyle x = \frac{-70}{4}$
$\displaystyle x = -17.5$

Why is the minimum value is wrong?
You need to use the positive value, ie:

$\displaystyle Sin(4x-20) = -1$
$\displaystyle 4x - 20 = 270$

Notice that $\displaystyle sin(270^o) = sin(-90^o)$

But your question specifically said "In each case, give the least positive value of x at which they occur."