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

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Answer 1 · 2017-04-06T21:38:41

See below.

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

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

How to solve this?f:RRf:RR\{2}->RR;f(x)=x^2/(x-2)->RR;f(x)=x^2/(x-2).Demonstrate that 8<=int_3^4f(x)dx<=98<=int_3^4f(x)dx<=9