Help With Homework: Step-by-Step Solutions Provided!

[wundumbbutt]

someone please help me with this homework. and show me the steps

http://img.photobucket.com/albums/v128/babycinnam0r0ll/scan.jpg

 

[whozum]

a) What trig function relates opposite and adjacent?
b) What trig function relates hypotenuse to either of the other sides? Theres two for this one.
c) Use the law of sines [itex] \frac{sin(a)}{A} = \frac{sin(b)}{B} = \frac{sin(c)}{C}[/tex]
d) Label the point with the right triangle D. Then $sin(10) = \frac{\overline{CD}}{14}$. Do something similar to find AD, then sum AD and DC to find AC.
e) Law of cosines. You have two angles, andyou knowthesum of all 3 is 180. Find the last angle, then apply the law of cosines.

 

[tacman]

Sure, I'd be happy to help you with your homework! First, let's take a look at the problem. It looks like you have a calculus question that involves finding the maximum and minimum values of a function. The function is given as f(x) = 3x^2 - 6x + 5.

To find the maximum and minimum values of a function, we need to use the first derivative test. This means we need to take the derivative of the function and set it equal to 0 to find the critical points. The critical points are where the function changes from increasing to decreasing or vice versa, and these points will give us the maximum and minimum values.

So, let's start by taking the derivative of f(x) using the power rule. The derivative of 3x^2 is 6x, the derivative of -6x is -6, and the derivative of 5 is 0. Therefore, the derivative of f(x) is 6x - 6.

Next, we set the derivative equal to 0 and solve for x.

6x - 6 = 0
6x = 6
x = 1

So, x = 1 is our critical point.

Now, we need to determine if this critical point is a maximum or a minimum. To do this, we use the second derivative test. This involves taking the second derivative of the function and plugging in our critical point. If the second derivative is positive, then the critical point is a minimum. If the second derivative is negative, then the critical point is a maximum.

The second derivative of f(x) is 6, so when we plug in x = 1, we get 6. Since the second derivative is positive, we know that x = 1 is a minimum.

To find the minimum value, we plug x = 1 back into the original function.

f(1) = 3(1)^2 - 6(1) + 5
f(1) = 2

Therefore, the minimum value of the function is 2, and it occurs at x = 1.

To find the maximum value, we can use the same process. However, instead of using the second derivative test, we can just look at the end behavior of the function. Since the coefficient of the x^2 term is positive, the end behavior of the