Quick calc hw question

MaRiNe23

United States747 Posts

April 15 2008 04:47 GMT

1. Let f be the function that is defined or all real numbers x and that has the following properties.

(i) f''(x)=24x-18
(ii) f'(1)=-6
(iii) f(2)=0

Find each x such that the line tangent to the graph of f at (x,f(x)) is horizontal.

2. Let f be the function defined by f(x)=ln(2+sinx) for pi<x<2pi.

(a) Find the absolute maximum value and the absolute minimum value of f. Show the analysis that leads to your conclustion.

(b) Find the x coordinate of each inflection point on the graph of f. Justify your answer.

Blah I'm so bad at calculus. I seriously don't do any homework or even pay attention in that class. Can't even start on most of my homework cuz I don't even understand wtf is going on, espcially word problems are just T_T for me..I honestly don't know why I'm even in it :/

xhuwin

United States476 Posts

April 15 2008 05:05 GMT

1.
We want to find the x coordinates of points (x, f(x)) such that the tangent line is horizontal. That basically means the derivative = f'(x) = 0 at these points, but we don't have a definition of f'(x).

How do we find f'(x)? We integrate f''(x) = 24x - 19 -> f'(x) = 12x^2 - 18x + c. To find c, we use the fact that f'(1) = -6. Then, knowing f'(x), the rest should be fairly straightforward.

PM me if you still need help.

Gyabo

United States329 Posts

April 15 2008 05:05 GMT

On April 15 2008 13:47 MaRiNe23 wrote:
Find each x such that the line tangent to the graph of f at (x,f(x)) is horizontal.

This means the first derivative of f at the point (x,f(x)) is 0 because the first derivative is the slope of the function and a horizontal line means 0 slope. (ie. f '(x) = 0)

First take the indefinite integral of (i). Don't forget the "+ C" at the end. Then use the initial condition (ii) to solve for C. Then plug in 0 for f '(x) and solve for x.

sigma_x

Australia285 Posts

April 15 2008 06:59 GMT

2a) Graphing should make the answer fairly obvious. Do this by first graphing 2+sinx, then applying the function "ln" to it.

For a more rigorous solution, consider f'(x)=cosx/(2+sinx) in the interval (pi, 2pi). Since -1<sinx<0 in this interval, then 0<1<2+sinx<2 so the denominator of f'(x) is positive, so it is suffice to note that:
1) (pi,3pi/2), cosx<0 ==> f'(x)<0
2) (3pi/2,2pi), cosx>0 ==> f'(x)>0 and;
3) x=3pi/2, cosx=0 ==> f'(x)=0.

This means that the absolute minimum in this interval must occur at x=3pi/2 with f(x)=0, and the maximum occurring at either x=pi or x=2pi. It is now trivial to show that these two values are in fact the same with f(x)=ln2. (Note: since the interval is open, these are not strictly their maximum, but rather their lowest upper bound).

Part 2b, use the identity (cosx)^2+(sinx)^2=1 and don't forget the interval.

wanderer

United States641 Posts

April 15 2008 07:45 GMT

I was doing the first problem for myself, and I don't see how the third piece of information is supposed to work into it all. Anyone wanna clarify?

sigma_x

Australia285 Posts

April 15 2008 11:06 GMT

On April 15 2008 16:45 wanderer wrote:
I was doing the first problem for myself, and I don't see how the third piece of information is supposed to work into it all. Anyone wanna clarify?

I don't think it's entirely relevant, though maybe it's implicit in the question to work out f(x).

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