Question

What is the absolute extreme of \[f(x) = \dfrac{x}{{{x^2} + 25}}\]on the interval \[[0,9]\]?

Answer 1

Hint: In this question we have to evaluate the absolute extrema of the function at the critical points and the end points of the interval. Absolute extrema are the largest and smallest the function will ever be and these four points represent the only places in the interval where the absolute extrema can occur.
Quotient rule of derivative:
\[{\left( {\dfrac{u}{v}} \right)’} = \dfrac{{u'v - uv'}}{{{v^2}}}\]

Complete step-by-step solution:
It is given that \[f(x) = \dfrac{x}{{{x^2} + 25}}\].
We need to find out the absolute extreme of \[f(x) = \dfrac{x}{{{x^2} + 25}}\] on the interval \[[0,9]\].
Let,\[u = x\& v = {x^2} + 25\]
Differentiating both sides with respect to x we get,
\[u' = \dfrac{{du}}{{dx}} = 1\]
\[v' = \dfrac{{dv}}{{dx}} = 2x\]
Now we can write,\[f(x) = \dfrac{u}{v}\]
We know the quotient rule of derivative is
\[{\left( {\dfrac{u}{v}} \right)’} = \dfrac{{u'v - uv'}}{{{v^2}}}\]
Using the quotient rule differentiation \[f(x) = \dfrac{x}{{{x^2} + 25}}\] both sides again with respect to x,
\[f'(x) = {\left( {\dfrac{u}{v}} \right)’} = \dfrac{{u'v - uv'}}{{{v^2}}} = \dfrac{{1\left( {{x^2} + 25} \right) - x \times 2x}}{{{{\left( {{x^2} + 25} \right)}^2}}}\]
\[\Rightarrow f'(x) = \dfrac{{{x^2} + 25 - 2{x^2}}}{{{{\left( {{x^2} + 25} \right)}^2}}}\]
\[\Rightarrow f'(x) = \dfrac{{ - {x^2} + 25}}{{{{\left( {{x^2} + 25} \right)}^2}}}\]
Now when we need to find any relative maximum or minimum, we need to set \[f'(x) = 0\].
So, to find the critical points we only need to make \[ - {x^2} + 25 = 0\], since \[{\left( {{x^2} + 25} \right)^2} \times 0 = 0\]
 Thus \[{x^2} = 25\]
\[\Rightarrow x = \pm 5\]
Therefore the critical points are \[x = \pm 5\].
 Now we need to find the value of the function at critical points and the end points.
\[f(0) = 0\]
\[f(9) = \dfrac{9}{{{9^2} + 25}} = \dfrac{9}{{81 + 25}} = \dfrac{9}{{106}} \simeq 0.085\]
Since the given interval is \[[0,9]\] we will consider only the critical value \[5\].
\[f(5) = \dfrac{5}{{{5^2} + 25}} = \dfrac{5}{{25 + 25}} = \dfrac{5}{{50}} = \dfrac{1}{{10}} = 0.1\]
The largest value is at \[f(5)\]and the smallest value is at \[f(0)\].
Hence the absolute maxima is at \[\left( {5,\dfrac{1}{{10}}} \right)\] and the absolute minima is at \[\left( {0,0} \right)\]
Note: An absolute maximum point is a point where the function obtains its greatest possible value. Similarly, an absolute minimum point is a point where the function obtains its least possible value.
To determine the absolute extrema of a function on a closed interval \[[a,b]\] we can apply the following steps,
1)Find all the critical points c on the open interval \[(a,b)\].
For that we need to make the first derivative of the function equals to zero and find out the values of x.