Thread: addition of ordinates

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

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.

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)

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).

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

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).

oh is it -0.306

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.

thanks you gun

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

thankyou heaps