
Maximum value of the function \[f\left( x \right) = \dfrac{x}{8} + \dfrac{2}{x}\] on the interval [1,6] is
A. $ 1 $
B. $ \dfrac{{17}}{8} $
C. $ \dfrac{{13}}{{12}} $
D. $ \dfrac{{17}}{8} $
Answer
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Hint: To answer the maximum value of the given function in the given interval can be calculated by some procedure like first find the derivative of f(x) then equate it to zero. Find the value of x for which their derivative is equal to zero. Then check the interval in which the first derivative of the function is increasing and decreasing. The interval in which the first derivative is decreasing will give the maximum value of the function.
Complete step-by-step answer:
Given function
\[f\left( x \right) = \dfrac{x}{8} + \dfrac{2}{x}\]
Now finding the first derivative of the above function.
We get,
\[f\prime \left( x \right) = \dfrac{1}{8} - \dfrac{2}{{{x^2}}} = \dfrac{{{x^2} - 16}}{{8{x^2}}} = 0\]
For maximum or minimum, f′(x) must vanish.
\[f\prime \left( x \right) = 0\]
So,
\[\dfrac{{{x^2} - 16}}{{8{x^2}}} = 0\]
We get for \[x = 4, - 4\]
In this question it is given the interval of x is
\[x \in \left[ {1,6} \right]\]
So $ x \ne - 4 $
Because in the given interval range of it only allows positive values of x not negative values of x.
Also, in \[\left[ {1,4} \right],f\prime \left( x \right) < 0 \Rightarrow f\left( x \right)\;\] is decreasing.
In \[\left[ {4,6} \right],f\prime \left( x \right) > 0 \Rightarrow f\left( x \right)\;\] is increasing.
\[\Rightarrow f\left( 1 \right) = \dfrac{{17}}{8}\]
\[\Rightarrow f\left( 6 \right) = \dfrac{{6}}{8} + \dfrac{2}{6} = \dfrac{3}{4} + \dfrac{1}{3} = \dfrac{{13}}{{12}}\]
Hence, maximum value of f(x) in [1,6] is \[\dfrac{{17}}{8}\]
So, the correct answer is “\[\dfrac{{17}}{8}\]”.
Note: To find the maximum value of the given function in the given interval we can use an another approach like finding the second derivative and check whether which value of x gives the second derivative negative corresponding to that value of x we will get the maximum value of the given function in the given interval.
Complete step-by-step answer:
Given function
\[f\left( x \right) = \dfrac{x}{8} + \dfrac{2}{x}\]
Now finding the first derivative of the above function.
We get,
\[f\prime \left( x \right) = \dfrac{1}{8} - \dfrac{2}{{{x^2}}} = \dfrac{{{x^2} - 16}}{{8{x^2}}} = 0\]
For maximum or minimum, f′(x) must vanish.
\[f\prime \left( x \right) = 0\]
So,
\[\dfrac{{{x^2} - 16}}{{8{x^2}}} = 0\]
We get for \[x = 4, - 4\]
In this question it is given the interval of x is
\[x \in \left[ {1,6} \right]\]
So $ x \ne - 4 $
Because in the given interval range of it only allows positive values of x not negative values of x.
Also, in \[\left[ {1,4} \right],f\prime \left( x \right) < 0 \Rightarrow f\left( x \right)\;\] is decreasing.
In \[\left[ {4,6} \right],f\prime \left( x \right) > 0 \Rightarrow f\left( x \right)\;\] is increasing.
\[\Rightarrow f\left( 1 \right) = \dfrac{{17}}{8}\]
\[\Rightarrow f\left( 6 \right) = \dfrac{{6}}{8} + \dfrac{2}{6} = \dfrac{3}{4} + \dfrac{1}{3} = \dfrac{{13}}{{12}}\]
Hence, maximum value of f(x) in [1,6] is \[\dfrac{{17}}{8}\]
So, the correct answer is “\[\dfrac{{17}}{8}\]”.
Note: To find the maximum value of the given function in the given interval we can use an another approach like finding the second derivative and check whether which value of x gives the second derivative negative corresponding to that value of x we will get the maximum value of the given function in the given interval.
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