IIT JEE , AIEEE Forum - Mathematics , Algebra - don't know wat to write....just help me !!.....i'll surely rate

netkid07

find max and min value of

(x^2+14x+9)/(x^2+2x+9)

this q is from quadratic eqns but i wanted to ask if this can be soved by calculus..............anyways plz solve it

n plz don't make fun of me if it's too easy coz i don't go coaching classes

thanking u in advance !!

Greatdreams

(x^2+14x+9)/(x^2+2x+9) = y

So cross multiplying,

x²(1 - y) + 2x( 7 - y) + 3(3 - y) = 0

Now if x is real ,

b² - 4 ac

0

So 4(7 - y)2 - 12 (1 - y)(3 - y)

0

49 - 14y + y² - 3( 3 - 4y + y2)

0

So 2 (y² + y - 20)

0

Thus (y + 5)(y - 4)

0

So y lies between -5 and 4

So maximum value is 4 and min is -5

rishabh29288

minimum value- 1
maximum value- 17/5
this is how i did it
write the whole thing as
1 + 12/x+ 2 + 9/x
now the whole thing will be minimum when the denom of the second part of the relation shall be maximum and that happens for x= -3 so u get 1 as the minimum
and the whole thing wil be max if the denom is min. which will happen when x+ 9/x is minimum considering only positive values
using a.m. > g.m
the minimum value fr x + 9/x will be 3

rishabh29288

yes its quite strange that thru the normal procedure as done by gr8dreams we get other value but i don't understand the mistake in my method

netkid07

can't anyone solve it by calculus or am gm inequality !!

joyfrancis

x²+14x+9 / x²+2x+9

= 1 + 12x/x²+2x+9

Now , just find maxima and minima of 12x/x²+2x+9

Let f(x)=12x/x²+2x+9

so f'(x) would come out to be = 108-12x²/(x²+2x+9)² .....by quotient rule

since x²+2x+9 is always > 0 the critical points would be only where

108-12x²=0

i.e x = plusminus3

at x = 3;12x/x²+2x+9 = 3/2

at x = -3;12x/x²+2x+9 = -3

So maximum value of x²+14x+9 / x²+2x+9 = 3/2+1=5/2

& min. value = -3+1=-2

joyfrancis

Here is the graph of the function.

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