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

2. 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.

3. 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)

4. 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).

5. 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

6. h(1) is still wrong.

Originally Posted by emakarov
the left-hand side is h(1), not h(x) for some indeterminate x.
Originally Posted by emakarov
h(1) is a number, it can't equal (1,-1), which is a pair of numbers.
Originally Posted by emakarov
Calculate... h(x) for x close to -1 (approximately).

7. oh is it -0.306

8. 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.

9. thanks you gun

10. Also, the maximum will be approximately at x = 0.48.

11. thankyou heaps