(Headbang)(Headbang)(Headbang)y=x^3 and Z=loge(x+1). Hence, using addition of ordinates, sketch the grph of w=z-y. use domain of (-1,-2].

please i need help with this what would be the new set of points. i dont get it

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- May 30th 2011, 01:32 AMBarneyaddition of ordinates
(Headbang)(Headbang)(Headbang)y=x^3 and Z=loge(x+1). Hence, using addition of ordinates, sketch the grph of w=z-y. use domain of (-1,-2].

please i need help with this what would be the new set of points. i dont get it - May 30th 2011, 02:04 AMemakarov
You probably mean the segment (-1,2]. (Why do we have to guess?)

Draw both graphs. Note approximately their intersection points. The difference will be zero in these points. Determine where the difference w is positive and negative. Calculate the value (approximately) of w for x = 0, 1, 2 and very close to -1.

In general, if you have two points (x, z) and (x, y), then the difference will obviously be (x, z - y). Determine this difference (at least visually and approximately) for any additional x coordinates that you'd like, and then smoothly connect the obtained points of the graph. - May 30th 2011, 02:21 AMBarney
hi thankyou so is this what i do to get the points:

h(x)=loge(1+1)-1^3 = (1,-1)

h(x)=loge(2+1)-2^3 = (2,-6.90)

h(x)=loge(0.1+1)0.1^3 = (0.1,0.094) - May 30th 2011, 02:35 AMemakarov
You need to define h since this is the first time it appears. Let h(x) = z(x) - y(x).

h(1) = log(1+1) - 1^3; the left-hand side is h(1), not h(x) for some indeterminate x.

h(1) is a number, it can't equal (1,-1), which is a pair of numbers.

log(2) - 1 is not -1.

Calculate h(0) and h(x) for x close to -1 (approximately). - May 30th 2011, 02:53 AMBarney
h(x)=loge(1+1)-1^3 = (1,1.69)

h(x)=loge(2+1)-2^3 = (2,-6.90)

h(x)=loge(0.1+1)0.1^3 = (0.1,0.094)

h(0)=0 - May 30th 2011, 03:25 AMemakarov
- May 30th 2011, 03:28 AMBarney
oh is it -0.306

- May 30th 2011, 03:31 AMemakarov
After you've done all of the above (note: I did not quote myself automatically), connect the points. You can calculate h(x) for other x's in the same way.

- May 30th 2011, 03:33 AMBarney
thanks you gun

- May 30th 2011, 03:34 AMemakarov
Also, the maximum will be approximately at x = 0.48.

- May 30th 2011, 03:39 AMBarney
thankyou heaps