How to solve this?#f:RR#\{2}#->RR;f(x)=x^2/(x-2)#.Demonstrate that #8<=int_3^4f(x)dx<=9#

1 Answer
Apr 6, 2017

See below.

Explanation:

We have

#f'(x)=(x^2-4x)/(x-2)^2#

and

#f'(x)# is monotonic increasing for #3 le x le 4#

with

#m_3=f'(3)=-3#
#m_4=f'(4)=0#

so

for #3 le x le 4->{(l_i(x)=f(3)+m_3(x-3) le f(x)),(l_s(x)=f(3)+m_4(x-3) ge f(x)):}#

but also defining #l_1(x) = f(4)+m_4(x-3)# and calculating

#x_m= l_i(x) nn l_1(x)=10/3#

we have that #l_2(x) = {(l_i(x), 3 le x le x_m),(l_1(x), x_m lt x le 4):}#

is such that

#l_2(x) le f(x)# for #3 le x le 4# so

#int_3^4 l_2(x)dx le int_3^4f(x)dx le int_3^4l_s(x)dx#

but

#int_3^4 l_2(x)dx=49/6# and
#int_3^4 l_s(x)dx=9#

so finally

#8 < 49/6 le int_3^4f(x)dx le 9#