given that f(x)=((mx^2 +4(√3)x+n)/(x^2 +1)) and f max=7 f min=−1 find the value of m and n


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Question Number 133179 by abdullahquwatan last updated on 19/Feb/21

$given that f (x) = \frac{{mx}^{2} + 4 \sqrt{3} x + n}{x^{2} + 1} and f \max = 7$ $f \min = - 1 find the value of m and n$
Commented by abdullahquwatan last updated on 20/Feb/21

$thx$
	Answered by EDWIN88 last updated on 21/Feb/21
	$Given f(x)=((mx^2 +4(√3)x+n)/(x^2 +1)) ; have minimum = −1 and maximum = 7. Find m and n. ((df(x))/dx) = (((2mx+4(√3))(x^2 +1)−2x(mx^2 +4(√3)x+n))/((x^2 +1)^2 ))=0 ⇒2mx^3 +4(√3)x^2 +2mx+4(√3)−2mx^3 −8(√3)x^2 −2nx = 0 −4(√3)x^2 +(2m−2n)x+4(√3) = 0 it follows that have 2 roots say it { ((x_1 ⇒for minimum value)),((x_2 ⇒for maximum value)) :} consider f(x_1 )=((mx_1 ^2 +4(√3)x_1 +n)/(x_1 ^2 +1)) = −1 ⇒ (m+1)x_1 ^2 +4(√3)x_1 +(n+1)=0 take discriminant = 0 ⇒ 12 = (m+1)(n+1) ; mn+m+n=11...(i) consider f(x_2 )=((mx_2 ^2 +4(√3)x_2 +n)/(x_2 ^2 +1))= 7 ⇒(m−7)x_2 ^2 +4(√3)x_2 +(n−7) = 0 take discriminant = 0 ⇒12 = (n−7)(m−7); mn−7(m+n)=−37...(ii) solving for m & n (i)−(ii)⇒ 8(m+n)=48 ; m+n = 6 substitute to eq(i) →mn+6 = 11 m(6−m)=5 ⇒m^2 −6m+5 = 0 ⇒(m−1)(m−5)=0→ { ((m=1,n=5)),((m=5,n=1)) :} then f(x)=((x^2 +4(√3)x+5)/(x^2 +1)) or f(x)=((5x^2 +4(√3)x+1)/(x^2 +1))$
	$Given f (x) = \frac{{mx}^{2} + 4 \sqrt{3} x + n}{x^{2} + 1}; have$ $minimum = - 1 and maximum = 7 . Find$ $m and n .$ $\frac{df (x)}{dx} = \frac{(2 mx + 4 \sqrt{3}) (x^{2} + 1) - 2 x ({mx}^{2} + 4 \sqrt{3} x + n)}{{(x^{2} + 1)}^{2}} = 0$ $\Rightarrow {2 mx}^{3} + 4 \sqrt{3} x^{2} + 2 mx + 4 \sqrt{3} - {2 mx}^{3} - 8 \sqrt{3} x^{2} - 2 nx = 0$ $- 4 \sqrt{3} x^{2} + (2 m - 2 n) x + 4 \sqrt{3} = 0 it follows that$ $have 2 roots say it {\begin{cases} x_{1} \Rightarrow for minimum value \\ x_{2} \Rightarrow for maximum value \end{cases}$ $consider f (x_{1}) = \frac{{mx}_{1}^{2} + 4 \sqrt{3} x_{1} + n}{x_{1}^{2} + 1} = - 1$ $\Rightarrow (m + 1) x_{1}^{2} + 4 \sqrt{3} x_{1} + (n + 1) = 0$ $take discriminant = 0$ $\Rightarrow 12 = (m + 1) (n + 1); mn + m + n = 11 . . . (i)$ $consider f (x_{2}) = \frac{{mx}_{2}^{2} + 4 \sqrt{3} x_{2} + n}{x_{2}^{2} + 1} = 7$ $\Rightarrow (m - 7) x_{2}^{2} + 4 \sqrt{3} x_{2} + (n - 7) = 0$ $take discriminant = 0$ $\Rightarrow 12 = (n - 7) (m - 7); mn - 7 (m + n) = - 37 . . . (ii)$ $solving for m & n$ $(i) - (ii) \Rightarrow 8 (m + n) = 48; m + n = 6$ $substitute to eq (i) \to mn + 6 = 11$ $m (6 - m) = 5 \Rightarrow m^{2} - 6 m + 5 = 0$ $\Rightarrow (m - 1) (m - 5) = 0 \to {\begin{cases} m = 1, n = 5 \\ m = 5, n = 1 \end{cases}$ $then f (x) = \frac{x^{2} + 4 \sqrt{3} x + 5}{x^{2} + 1} or f (x) = \frac{{5 x}^{2} + 4 \sqrt{3} x + 1}{x^{2} + 1}$
	Commented by bramlexs22 last updated on 21/Feb/21

	Commented by bramlexs22 last updated on 21/Feb/21

	$yes correct sir$
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