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)&x=3 so,the monotonicity of f(x) is f(n)<f((1/2))_(∣n<(1/2)) ,f((1/2))>f(2),f(2)<f(p)_(∣p>2.) Q_1 has been proved finished Q_(2 ) ,set ax^2 −(a+6)x+3ln x>−6 when x∈[2,3e], make the f(x) dive two part as g(x)=ax^2 −(6+a)x&k(x)=−3(ln x+2), the middle value of g(x) is x=((a+6)/(2a)) when x=3e, k_(min) (3e)=−15, x=2, k_(max) (2)=−3(ln 2+2) g(2)=2a−12>k_(max) (x)⇒a>3−(3/2)ln 2 when x=(1/2)+(3/a), g(x)=−(((a+6)^2 )/(4a))>−3(ln (((a+6)/(2a)))+2)&a>0&2<(1/2)+(3/a)<3e


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

$a s w e c a n k n o w f o r Q_{1}, s e t t h e f u n v c t i o n i n q u e s t i o n a s f (x)$ $a n d m a k e t h e f i r s t s t e p a s :$ $a = 1, f (x) = x^{2} - 7 x + 3 \ln x$ $d f (x) = 2 x - 7 + \frac{3}{x} = h (x)$ $s e t h (x) = 0 \Rightarrow x = \frac{1}{2} & x = 3$ $Missing \left or extra \right$ $Q_{1} h a s b e e n p r o v e d f i n i s h e d$ $Q_{2}, s e t {a x}^{2} - (a + 6) x + 3 \ln x > - 6 w h e n x \in [2, 3 e],$ $m a k e t h e f (x) d i v e t w o p a r t a s$ $g (x) = {a x}^{2} - (6 + a) x & k (x) = - 3 (\ln x + 2),$ $t h e m i d d l e v a l u e o f g (x) i s x = \frac{a + 6}{2 a}$ $w h e n x = 3 e, k_{m i n} (3 e) = - 15, x = 2, k_{m a x} (2) = - 3 (\ln 2 + 2)$ $g (2) = 2 a - 12 > k_{m a x} (x) \Rightarrow a > 3 - \frac{3}{2} \ln 2$ $w h e n x = \frac{1}{2} + \frac{3}{a}, g (x) = - \frac{{(a + 6)}^{2}}{4 a} > - 3 (\ln (\frac{a + 6}{2 a}) + 2) & a > 0 & 2 < \frac{1}{2} + \frac{3}{a} < 3 e$
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