find the minimum and maximum value of the quadratic functions a) 4x^2 + 5x + 1 b) x + (2/x) = 3 c) x^2 − (x/4) + 6 hence draw each draw


Question and Answers Forum

All Questions Topic List
Others Questions
Previous in All Question Next in All Question
Previous in Others Next in Others
Question Number 39607 by Rio Mike last updated on 08/Jul/18

$f i n d t h e$ $m i n i m u m a n d m a x i m u m v a l u e$ $o f t h e q u a d r a t i c f u n c t i o n s$ $a) 4 x^{2} + 5 x + 1$ $b) x + \frac{2}{x} = 3$ $c) x^{2} - \frac{x}{4} + 6$ $h e n c e d r a w e a c h d r a w$
	Answered by tanmay.chaudhury50@gmail.com last updated on 08/Jul/18
	$a)4x^2 +5x+1 4(x^2 +(5/4)x+(1/4)) 4{(x^2 +2.x.(5/8)+((25)/(64))+(1/4)−((25)/(64)))} 4{(x+(5/8))^2 +((16−25)/(64))} 4{(x+(5/8))^2 +((−9)/(64))} 4(x+(5/8))^2 −(9/(16)) (x+(5/8))^2 >0 as the value of x increases so the value of (x+(5/8))^2 increases...hence 4x^2 +5x+1 has no maximum value..its minimum value is ((−9)/(16)) when x=((−5)/8) other method... by using calculus y=4x^2 +5x+1 (dy/dx)=8x+5 for min/max (dy/dx)=0 so x=−(5/8) (d^2 y/dx^2 )=8 that means (d^2 y/dx^2 )>0 so 4x^2 +5x+1 has minimum value at x=−(5/8) 4.(((25)/(64)))+5(((−5)/8))+1 ((25)/(16))−((25)/8)+1 ((25−50+16)/(16))=((−9)/(16))$
	$a) 4 x^{2} + 5 x + 1$ $4 (x^{2} + \frac{5}{4} x + \frac{1}{4})$ $4 {(x^{2} + 2 . x . \frac{5}{8} + \frac{25}{64} + \frac{1}{4} - \frac{25}{64})}$ $4 {{(x + \frac{5}{8})}^{2} + \frac{16 - 25}{64}}$ $4 {{(x + \frac{5}{8})}^{2} + \frac{- 9}{64}}$ $4 {(x + \frac{5}{8})}^{2} - \frac{9}{16}$ ${(x + \frac{5}{8})}^{2} > 0$ $a s t h e v a l u e o f x i n c r e a s e s s o t h e v a l u e o f$ ${(x + \frac{5}{8})}^{2} i n c r e a s e s . . . h e n c e 4 x^{2} + 5 x + 1 h a s n o$ $m a x i m u m v a l u e . . i t s m i n i m u m v a l u e i s$ $\frac{- 9}{16} w h e n x = \frac{- 5}{8}$ $o t h e r m e t h o d . . .$ $b y u s i n g c a l c u l u s$ $y = 4 x^{2} + 5 x + 1$ $\frac{d y}{d x} = 8 x + 5$ $f o r m i n / m a x \frac{d y}{d x} = 0 s o x = - \frac{5}{8}$ $\frac{d^{2} y}{{d x}^{2}} = 8 t h a t m e a n s \frac{d^{2} y}{{d x}^{2}} > 0$ $s o 4 x^{2} + 5 x + 1 h a s m i n i m u m v a l u e a t x = - \frac{5}{8}$ $4 . (\frac{25}{64}) + 5 (\frac{- 5}{8}) + 1$ $\frac{25}{16} - \frac{25}{8} + 1$ $\frac{25 - 50 + 16}{16} = \frac{- 9}{16}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 08/Jul/18
	$c)y=((4x^2 −x+24)/4) (1/4){(2x)^2 −2.2x.(1/4)+(1/(16))+24−(1/(16))} (1/4){(2x−(1/4))^2 +((24×16−1)/(16))} as the value of x increases so the value of expression increases..so the expression has no maximum value.. its minimum value is ((24×16−1)/(4×16)) when x=(1/8) min value=((383)/(64)) when x=(1/8) by calculus y=x^2 −(x/4)+6 (dy/dx)=2x−(1/4) for min/max (dy/dx)=0 so x=(1/8) (d^2 y/dx^2 )=2 so (d^2 y/dx^2 )>0 so the expression has min value at x=(1/8) min valud ((1/8))^2 −(1/(32))+6 ((1−2+384)/(64)) =((383)/(64))$
	$c) y = \frac{4 x^{2} - x + 24}{4}$ $\frac{1}{4} {{(2 x)}^{2} - 2.2 x . \frac{1}{4} + \frac{1}{16} + 24 - \frac{1}{16}}$ $\frac{1}{4} {{(2 x - \frac{1}{4})}^{2} + \frac{24 \times 16 - 1}{16}}$ $a s t h e v a l u e o f x i n c r e a s e s s o t h e v a l u e o f$ $e x p r e s s i o n i n c r e a s e s . . s o t h e e x p r e s s i o n h a s$ $n o m a x i m u m v a l u e . .$ $i t s m i n i m u m v a l u e i s \frac{24 \times 16 - 1}{4 \times 16} w h e n x = \frac{1}{8}$ $m i n v a l u e = \frac{383}{64} w h e n x = \frac{1}{8}$ $b y c a l c u l u s$ $y = x^{2} - \frac{x}{4} + 6$ $\frac{d y}{d x} = 2 x - \frac{1}{4}$ $f o r m i n / m a x \frac{d y}{d x} = 0 s o x = \frac{1}{8}$ $\frac{d^{2} y}{{d x}^{2}} = 2 s o \frac{d^{2} y}{{d x}^{2}} > 0$ $s o t h e e x p r e s s i o n h a s m i n v a l u e a t x = \frac{1}{8}$ $m i n v a l u d {(\frac{1}{8})}^{2} - \frac{1}{32} + 6$ $\frac{1 - 2 + 384}{64} = \frac{383}{64}$
	Answered by MJS last updated on 08/Jul/18

	$y = {a x}^{2} + b x + c$ $z e r o s a t x = - \frac{b}{2 a} \pm \frac{\sqrt{b^{2} - 4 a c}}{2 a}$ $a > 0 \Rightarrow m i n a t x = - \frac{b}{2 a}; y = \frac{- b^{2} + 4 a c}{4 a^{2}}$ $a < 0 \Rightarrow m a x a t x = - \frac{b}{2 a}; y = \frac{- b^{2} + 4 a c}{4 a^{2}}$ $y = x^{2} + p x + q \Rightarrow m i n a t x = - \frac{p}{2}; y = \frac{- p^{2} + 4 q}{4}$ $z e r o s a t x = - \frac{p}{2} \pm \frac{\sqrt{p^{2} - 4 q}}{2}$ $y = - x^{2} + p x + q \Rightarrow m a x a t x = \frac{p}{2}; y = \frac{p^{2} + 4 q}{4}$ $z e r o s a t x = \frac{p}{2} \pm \frac{\sqrt{p^{2} + 4 q}}{2}$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum