Algebra> let f(x) = [ ln ( 7x-x 2 ) ] 3/2 , then 1...

let f(x) = [ ln ( 7x-x2) ]3/2 , then12f is defined only on R+and is stricly increasing. f is defined on an interval of finite length and is strictly increasing. f is defined on an interval of finite length and is bounded. range of f contains 1.

tmk , 8 Years ago

Grade 12

2 Answers

Vikas TU

Last Activity: 8 Years ago

1) 7x – x^2 should be strictly greater than zero for log property.

therfore, 7x – x^2 > 0

x belongs to (0,7)

Thus x is defined in this interval.

For strictly increasing and decreasing we check f’(x)

f’(x) = 3/2*[log(7x – x^2)]^1/2 * (7 – 2x)
NOW log term in underroot will always be positive
also 7 – 2x is > 0
for x lies in (0,7).

Hence strictly increasing.

2nd option. is the appropriate one.

Riddhish Bhalodia

Last Activity: 8 Years ago

lets check the domain
as the term inside squareroot must be >= 0
=> term inside ln be >=1
$7x-x^2 \geq 12 \quad \Rightarrow \quad (x-3)(x-4) \leq 0$
thus
$x \in [3,4]$
now at x=3 and x=4 ,f(3)= f(4) = 0 hence it cannot be strictly increasing
and it’s bounded.
the maximum value that f(x) can get is at x = 3.5
evaluate that
so now it’s solved