can anyone help me w/ this prob...someone explain it 2 me but i still dont get it...appreciate it?

Question

http://i181.photobucket.com/albums/x85/ctti_3/untitled-31.jpg

i think to make it look clearer you just click into it

Answer 1

You can solve it by taking the first derivative of each function, then solving for x=1.

It is obvious that the slope of the graph is upward at (1,1), so the first derivative of the function at x=1 must be positive.

Of the functions you listed, #4 is the only one where the first derivative is positive at x=1.

Taking the second derivative won't work. First, you cannot eyeball a point of inflection. Second, even if (1,1) were a point of inflection, then both #2 and #4 will have a second derivative of 0 at x=1. But #2 is wrong.

Answer 2

Hi
From the look of the graph you would say that (1,1) is a point where the curvature of the graph changes - thats a point where the graphs curve goes from pointing up to pointing down.
This point is a non-stationary point of inflection. At these points the second derivative will be 0.
So you just differentiate each of the functions twice - put x=1 into these second derivatives and see which give you values of 0.

In question 1
the second derivative f''(x) = -2 -2x
f''(1) = -2-2 =-4 so its not this one.

Could not read qu 2

In question 3
the second derivative f''(x) = -12x+2x
f''(1) = -12+2 =-10 so its not this one.

Question 4 looks good.
In question 4
the second derivative f''(x) = 2 -2x
f''(1) = 2-2 =0 so it could be this one.

Second last one isn't it, I can't read the last one.


I've probably used a sledgehammer on this one - other answers are simpler!!!!

Answer 3

Since f(x) will be very large and negative for large values of x, the coefficient of x^3 must be negative. That eliminates choices 2, 3, and 5. Inserting x = 1 into the three remaining functions yields f(1) = 1, 1, and 1, so that doesn't help. Let's see if the slope of all of these is positive at (1,1). The slopes are:
1. f '(x) = 3 -2x - x^2; f '(1) = 0. This isn't it.
4. f '(x) = 3 + 2x - x^2; f '(1) = 4. That's possible.
5. f '(x) = -3 + 2x + x^2; f '(1) = 0. This isn't it.

Looks like choice 4 is correct.

Answer 4

The point P(1,1) is a point of inflection and occurs only when the 2nd derivative is = 0.

Answer 4 is the only equation whose 2nd derivative is 0 when x =1, so it is the correct answer.

f(x) = -8/3 +3x +x^2-x^3/3
f'(x) = 3 +2x -x^2
f"(x) = 2-2x
f"(1) = 2-2*1 = 0

Answer 5

I wrote it: http://i11.tinypic.com/6cpkpjc.jpg

Explanation: You must determine its first derivative, then determine its second derivative. When second derivative=0, u can find X. Put these X in its equation, finally find Y.

Answer 6

I can't read it cause it didn't scan well. It looks like a cubic to me.