PLEASE Help me with this questions: limits, tangent lines, etc.

[maor]

#8. Find the limit: $\displaystyle \displaystyle \lim_{x\, \rightarrow\, 0}\,$$\displaystyle \dfrac{1\, -\, \cos^2(x)}{x}$

#16. Find an equation of the tangent line to the graph of $\displaystyle \, f(\theta)\,=\, \tan(\theta)\,$ at the point $\displaystyle \, \left(\dfrac{\pi}{4},\, 1\right).$

#17. Use a graphing utility to graph $\displaystyle \, f(x)\, =\, \dfrac{3x^2\, -\, 8}{x^2\, -\, 4}\,$ and its derivative, $\displaystyle \, f'(x),\,$ on the same coordinate axes. Then use the graph to describe the behavior of $\displaystyle \, f\,$ at that value of $\displaystyle \, x\,$ where $\displaystyle \, f'(x)\, =\, 0.$

#20. Let $\displaystyle \, q(x)\, =\, \dfrac{f(x)}{g(x)}.\,$ Use the figure to find $\displaystyle \, q'(5).$

 

[Deleted member 4993]

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

 

[maor]

What are your thoughts?

Please share your work with us ...even if you know it is wrong

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...Before-Posting

I don't know how to start.

 

[Harry_the_cat]

Q20. Since q(x) is a quotient, then the quotient rule would be a good place to start.

 

[Deleted member 4993]

I don't know how to start.

Following up on post-above:

Do you know the quotient rule of differentiation?

Also note that, f(x) = constant at x = 5.

 

[stapel]

I don't know how to start.

You have no idea how to get started on any of them? So you're not familiar with taking limits, let alone taking derivatives? :shock:

 

[Harry_the_cat]

Q16. To find the equation of any line, you need to know 2 things: a point on the line and the gradient of the line.

You are given the point. You need to calculate the gradient of the tangent at that point. (Hint: Derivative)

Once you have a point and the gradient, you can proceed to find the equation of the tangent. (Hine: A straight line has equation y=mx+c)

 

[Paul Belino]

Equation of a line

The question is unclear but maybe I can help:

To get the equation of a line y = mx + c there are two ways:
1: Knowing TWO points on the line and so the gradient between the points can be found
OR
2: Knowing ONE point on the line AND the gradient.

For method 2 m is already given. If the point on the line is say (a,b) then this is a solution to y=mx +c
You just put in a and b for x and Y resp in to the equation of the line.

So b = ma + c so c = (b-ma)

SO the equation becomes y = mx + (b-ma)
so y = m(x-a) +b

In method 1 let the two points be (a,b) and (e,f)

Then we have a triangle whose lengths of sides are (a-d) and (e-f) in the x and y directions resp.
You are right in saying that the gardient is also the tangent of the line.
So in our triangle the tangent is (e-f)/(a-d)
so m = (e-f)/(a-d)

Now that m is known you can proceed with method 2