• Oct 16th 2009, 08:13 PM
Gitano
How is it going guys? I have a test next week and there are a couple of questions on my test review that I don't understand. Here they are:

1. For what values of x does the graph f have a horizontal tangent.
a. f(x)= x + 2cosx b. f(x)= (e^x) (sinx)

- Regarding these problems, I know how to find the derivative, but have no idea what to do next to solve it.

2. Let f(x)=x-2cosx, where x is greater than or equal to 0, but less than or equal to 2 pie.
a. On what intervals is f increasing or decreasing?
b. On what intervals is f concave upward or concave downward?

3. Find equations of both lines that are tangent to the curve y=1+x^3 and parallel to the line 12x-y=1.

Help or any feedback would be greatly appreciated, thanks.
• Oct 16th 2009, 08:25 PM
Quote:

Originally Posted by Gitano
How is it going guys? I have a test next week and there are a couple of questions on my test review that I don't understand. Here they are:

1. For what values of x does the graph f have a horizontal tangent.
a. f(x)= x + 2cosx b. f(x)= (e^x) (sinx)

- Regarding these problems, I know how to find the derivative, but have no idea what to do next to solve it.

When a line is horizontal, it's slope is zero. So the slope of the tangent lines are equal to zero at the points where the derivative is equal to zero. So just find the derivative:

(for your first example "a")

$\displaystyle f'(x)=1-2sin(x)$

Set to zero and solve.

$\displaystyle sin(x)=\frac{1}{2}$

$\displaystyle x=\frac{\pi}{6}$
• Oct 16th 2009, 08:32 PM
Gitano
That makes sense, thanks.

Anyone know how to solve problems 2 and 3 ?
• Oct 16th 2009, 08:34 PM
2. Let f(x)=x-2cosx, where x is greater than or equal to 0, but less than or equal to 2 pie.
a. On what intervals is f increasing or decreasing?
b. On what intervals is f concave upward or concave downward?

Ok, so we want examine the derivative of the function $\displaystyle f(x)=x-2cosx$ on the interval $\displaystyle [0,2\pi]$. If $\displaystyle f'(x)>0$, then the function is increasing. If $\displaystyle f'(x)<0$, then the function is decreasing.

If $\displaystyle f''(x)>0$, then the function is concave upward.
• Oct 16th 2009, 08:35 PM
3. Find equations of both lines that are tangent to the curve y=1+x^3 and parallel to the line 12x-y=1.

Do you know how to find the equation for a tangent to the curve?
• Oct 16th 2009, 08:53 PM
Gitano
Quote:

Originally Posted by adkinsjr
2. Let f(x)=x-2cosx, where x is greater than or equal to 0, but less than or equal to 2 pie.
a. On what intervals is f increasing or decreasing?
b. On what intervals is f concave upward or concave downward?

Ok, so we want examine the derivative of the function $\displaystyle f(x)=x-2cosx$ on the interval $\displaystyle [0,2\pi]$. If $\displaystyle f'(x)>0$, then the function is increasing. If $\displaystyle f'(x)<0$, then the function is decreasing.

If $\displaystyle f''(x)>0$, then the function is concave upward.

I know those rules, but I am not quite sure on how to solve it. The derivative would be 1+2sinx=0, then what?

Quote:

Originally Posted by adkinsjr
3. Find equations of both lines that are tangent to the curve y=1+x^3 and parallel to the line 12x-y=1.

Do you know how to find the equation for a tangent to the curve?

Only when there is a given point.
• Oct 16th 2009, 09:57 PM
Quote:

Originally Posted by Gitano
I know those rules, but I am not quite sure on how to solve it. The derivative would be 1+2sinx=0, then what?

Only when there is a given point.

All you have to do is solve the inequality and then right the answer in interval notation:

$\displaystyle 1+2sin(x) >0$

$\displaystyle 2sin(x)>-1$

$\displaystyle sin(x)>-\frac{1}{2}$

$\displaystyle Arcsin(-\frac{1}{2}) = \frac{7\pi}{6},\frac{11\pi}{6}$

Now just draw a graph of the sine function and plot the points on the curve. You'll see that the function is greater than $\displaystyle -\frac{1}{2}$ on the intervals: $\displaystyle sin(x)>-\frac{1}{2} on [0,\frac{7\pi}{6}) and (\frac{11\pi}{6},2\pi]$

Therefore, the function f is increasing on this interval.