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)xcolor(white)(aaaaaa)-oocolor(white)(aaaaaa)-3color(white)(aaaaaaaaaa)3color(white)(aaaaaaa)+oo

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

color(white)(aaaa)f(x)color(white)(aaaaaaaa)color(white)(aa)-0.09color(white)(aaaa)color(white)(a)1color(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.