basic calculus problems
my son just asked me to help him with some calculus but i haven't done calculus in 30 years so i can't really help him... i was hoping that someone here could help. his homework is do tomorrow. so, time is essential.. well, i was hoping that someone could hopefully show me how to do these problems so i could show my son how to do them.
Thanks in advance to all that this may concern.
philipsk is a dad he is entitled to our help to appear omnicient in the eyes
Originally Posted by ThePerfectHacker
of his chidren.
I should have added that he should not say that he is getting help to his son, I forgot to.
Originally Posted by CaptainBlack
2. Find the equation of the tangent to the curve at
The equation of a line (and the tangent is a line) is of the form , and this is a tangent at the given point on the curve if is equal to the slope or gradient of the curve at the point and c is chosen so that the line passes through the point.
Now the slope of the curve at a point is equal to the derivative of with respect to [tex][x/math] at the point. To evaluate this derivative we need the quotient rule:
with and , so , and
Now we could simplify this but I think it better to just evaluate it as it is at , which I make: .
So our line has the form: . Now when , so: , hence , so finally we have the equation of the tangent at the given point is:
(now I expect one of the other helpers will point out some dreadful mistake in my algebra:( )
3. Find the equation of the line tangent to at .
Same method as before.
At the point this is:
So we have the tangent line:
at the point . Again, we need b, so:
Thus the tangent line is:
Again, I have attached a graph. (The scale of the y-axis hides some of the features of the graph.)
For #7, polynomials are among the easiest functions to differentiate. They're everywhere continuous and very well-behaved.
As you can see, you have zeros at x=3 and x=-2
See the graph. Note the horizontal tangents as the maximum and minimum.