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# Help please and thank u! H.S. math!?

for the following, assume s(1)=-2, s'(1)=7, s"(1)=-1, t(1)=4, t'(1)=4, and t"(1)=-1.

a) let u(x)= s(x)/t(x). Then find u(1), u'(1), and u''(1).

b) is u(x) increasing or decreasing at x=1?

c) is u'(x) increasing or decreasing at x=1?

d) sketch a graph pf y=u(x) near x=1.

### 2 Answers

- johnLv 48 years agoFavorite Answer
u(x) = s(x)/t(x)

then

u'(x) = [s'(x)t(x) - s(x)t'(x)] / t(x)^2

u''(x) = {[s''(x)t(x)+s'(x)t'(x)-s'(x)t'(x)-s(x)t''(x)]t(x)^2 - [s'(x)t(x) - s(x)t'(x)](2)t(x)t'(x)}/t(x)^4

u''(x) = {[s''(x)t(x)-s(x)t''(x)]t(x)^2 - [s'(x)t(x) - s(x)t'(x)](2)t(x)t'(x)}/t(x)^4

a)

u(1) = s(1)/t(1) = -2 / 4 = -1/2

u'(1) = [s'(1)t(1) - s(1)t'(1)]/t(1)^2 =[7x4-(-2)4]/4^2 = 36/16 = 9/4

u''(1) = {[(-1)(4)-(-2)(-1))]4^2 - [(7)(4)-(-2)(4)](2)(4)(4)} / 4^4

u''(1) = {[-96]-[1152]} /4^4

u''(1) = - 39 /8

b) u'(1)=9/4 >> positive so is increasing

c) u''(1) = - 39/8 >> negative so u'(1) is decreasing

d) work it out yourself

cheers

- ajedrezLv 78 years ago
a) u(1) = -2/4 or -0.5

u'(x) = t(x)*s'(x) - s(x)*t'(x) over t(x) squared

4(1) - -2(4) over 16

4 + 8 over 16 = 3/4

u''x = t'(x)*s''(x) - s'(x)*t''(x) over t'(x) squared

u''(1) = 4(1) - 1(-1) over 4^2

4 + 1 over 16 = 5/16

They're both increasing because the slope is positive.

I hope this information was very helpful.