Please solve q 16?

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2 Answers
Jun 7, 2018

i) 1

Explanation:

The question is:
For all real values of #x#, the maximum value of the expression #x/(x^2-5x+9)# is -
i) 1; ii) 45; iii) 90; iv) None of these.

Could you type it in next time? The photo's a bit hard to read - not all of us have good eyesight. Edit: And in demonstration of this, I originally misread a sign in the question! Fixed now. Apologies for the error - it really isn't very easy for me to read.

To find the extrema of the function, set its derivative to 0.
#f(x)=x/(x^2-5x+9)#, so, by the quotient rule,
#f'(x)=((x^2-5x+9)-x(2x-5))/(x^2-5x+9)^2=(-x^2+9)/(x^2-5x+9)^2#

#f'(x)=0rArrx^2=9rArrx=+-3#

To characterise the two points found, consider the value of the second function derivative at these points.

By the quotient rule
#f''(x)=(-2x(x^2-5x+9)^2+(x^2-9)*2(x^2-5x+9)(2x-5))/(x^2-5x+9)^4#
#=(-2x(x^2-5x+9)+2(x^2-9)(2x-5))/(x^2-5x+9)^3#

So
#f''(+3)=(-6(9-15+9)+2(9-9)(6-5))/(9-15+9)^3=(-6*3+0)/3^3=-2/3<0#

So #x=3# is a function maximum.
Note that #f(3)=3/(9-15+9)=3/3=1#, which would match given answer (i).

Consider the other extremum now, at #x=-3#:
#f''(-3)=(+6(9+15+9)+2(9-9)(-6-5))/(9+15+9)^3=(+6*33+0)/33^3=2/363>0#

So #x=-3# is a function minimum and therefore not of interest as a possible answer to the question.

We now know that as #x->-oo#, the function value increases monotonically. It may increase to a limit, or it may increase to infinity. Consider the behaviour of the function as #x->-oo#:
the #x^2# term in the denominator increases in size much more rapidly than the #x# term in the numerator. So as #x->-oo#, #f(x)->0#.

Thus the answer is " i) 1 ", which maximum value for #f(x)# is reached at #x=3#.

Check the solution for sanity by plotting the graph of the function:
graph{x/(x^2-5x+9) [-8.89, 8.885, -4.444, 4.44]}

Jun 9, 2018

The answer is #option (1)#

Explanation:

The function is

#f(x)=x/(x^2-5x+9)#

The domain of #f(x)# is #RR# as the discriminant of the denominator is #Delta<0#

The derivative is calculated with the quotient rule.

#f'(x)=(1*(x^2-5x+9)-x(2x-5))/(x^2-5x+9)^2#

#=(x^2-5x+9-2x^2+5x)/(x^2-5x+9)^2#

#=(-x^2+9)/(x^2-5x+9)^2#

The critical points are when

#f'(x)=0#

#<=>#, #-x^2+9=0#

#<=>#, #x=+-3#

Build a variation chart

#color(white)(aaaa)##x##color(white)(aaaaaa)##-oo##color(white)(aaaaaa)##-3##color(white)(aaaaaaaaaa)##3##color(white)(aaaaaaa)##+oo#

#color(white)(aaaa)##"sign f'(x)"##color(white)(aaaa)##-##color(white)(aaaa)##0##color(white)(aaaa)##+##color(white)(aaaa)##0##color(white)(aaaa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaaa)##↘##color(white)(aa)##-0.09##color(white)(aaaa)##↗##color(white)(a)##1##color(white)(aaaa)##↘#

The maximum value is #=1#

The answer is #option (1)#

graph{x/(x^2-5x+9) [-7.9, 7.9, -3.95, 3.95]}

#2nd# way

The function is a maximum when the denominator is a minimum.

The denominator is

#x^2-5x+9=x^2-5x+25/4+9-25/4#

#=(x-5/2)^2-11/4#

The function is

#f(x)=x/((x-5/2)^2+11/4)#

is a maximum when

#x=5/2#

#=>#, #f(5/2)=(5/2)/(11/4)=5/2*4/11=10/11#

This is the maximum value.