as-we-can-know-for-Q-1-set-the-funvction-in-question-as-f-x-and-make-the-first-step-as-a-1-f-x-x-2-7x-3ln-x-df-x-2x-7-3-x-h-x-set-h-x-0-x-1-2-amp-x-3-so-the-monotonicity-of-f-x-is-f-n-l

Question Number 182657 by Strengthenchen last updated on 12/Dec/22

a s w e c a n k n o w f o r Q_{1}, s e t t h e f u n v c t i o n i n q u e s t i o n a s f (x)

a n d m a k e t h e f i r s t s t e p a s :

a = 1, f (x) = x^{2} - 7 x + 3 \ln x

d f (x) = 2 x - 7 + \frac{3}{x} = h (x)

s e t h (x) = 0 \Rightarrow x = \frac{1}{2} & x = 3

Missing \left or extra \right

Q_{1} h a s b e e n p r o v e d f i n i s h e d

Q_{2}, s e t {a x}^{2} - (a + 6) x + 3 \ln x > - 6 w h e n x \in [2, 3 e],

m a k e t h e f (x) d i v e t w o p a r t a s

g (x) = {a x}^{2} - (6 + a) x & k (x) = - 3 (\ln x + 2),

t h e m i d d l e v a l u e o f g (x) i s x = \frac{a + 6}{2 a}

w h e n x = 3 e, k_{m i n} (3 e) = - 15, x = 2, k_{m a x} (2) = - 3 (\ln 2 + 2)

g (2) = 2 a - 12 > k_{m a x} (x) \Rightarrow a > 3 - \frac{3}{2} \ln 2

w h e n x = \frac{1}{2} + \frac{3}{a}, g (x) = - \frac{{(a + 6)}^{2}}{4 a} > - 3 (\ln (\frac{a + 6}{2 a}) + 2) & a > 0 & 2 < \frac{1}{2} + \frac{3}{a} < 3 e