Question

Please solve q 16?

Answer 1

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]}

Answer 2

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.