Question

Let k and K be the minimum and maximum value of the function\[f(x) = \dfrac{{{{(1 + x)}^{0.6}}}}{{1 + {x^{0.6}}}}\]in \[[0,1]\] respectively, then the ordered pair (k, K) is equals to
A) \[({2^{ - 0.4}},1)\]
B) \[({2^{ - 0.6}},1)\]
C) \[({2^{ - 0.4}},{2^{0.6}})\]
D) \[(1,{2^{ - 0.6}})\]

Answer 1

Hint: As per the general concept in order to calculate the maxima and minima of any function differentiate the term given and equate it to zero and then obtain the value of x. Hence, that x will give us either maxima or minima value of the function. And also if either one value of x is calculated then calculate another value of x using the domain of x as per mentioned in the question.

Complete step by step solution: As the given \[f(x) = \dfrac{{{{(1 + x)}^{0.6}}}}{{1 + {x^{0.6}}}}\] in \[[0,1]\] is taken into consideration,
So, calculating the differentiation of the above-given function using division rule of differentiation.
As, \[d(\dfrac{u}{v}) = \dfrac{{v(u') - u(v')}}{{{v^2}}}\] so apply this condition in the above function as,
\[
  f'(x) = \dfrac{{(1 + {x^{0.6}})0.6{{(1 + x)}^{0.6 - 1}} - {{(1 + x)}^{0.6}}(0.6){x^{0.6 - 1}}}}{{{{(1 + {x^{0.6}})}^2}}} = 0 \\
   \Rightarrow \dfrac{{(1 + {x^{0.6}})0.6{{(1 + x)}^{ - 0.4}} - {{(1 + x)}^{0.6}}(0.6){x^{ - 0.4}}}}{{{{(1 + {x^{0.6}})}^2}}} = 0 \\
 \]
Hence, now solve the equation and calculate the value of x from the above obtained equation as,
\[ \Rightarrow (1 + {x^{0.6}}){(1 + x)^{ - 0.4}} - {(1 + x)^{0.6}}{x^{ - 0.4}} = 0\]
On rearranging we get,
\[ \Rightarrow \dfrac{{(1 + {x^{0.6}})}}{{{{(1 + x)}^{0.4}}}} = \dfrac{{{{(1 + x)}^{0.6}}}}{{{x^{0.4}}}}\]
Cross multiply and simplify the above terms and solve for x.
\[
   \Rightarrow ({x^{0.4}} + x) = {(1 + x)^{0.6}}{(1 + x)^{0.4}} \\
   \Rightarrow ({x^{0.4}} + x) = (1 + x) \\
   \Rightarrow {x^{0.4}} = 1 \\
 \]
Hence, we can conclude that the value of x is \[{\text{1}}\].
Graph of the given function:

Hence from the graph, we can conclude that the given function is decreasing in the given range so at \[x = 1\],
The minimum value of the function will be obtained and at \[x = 0\] maximum value of the function will be obtained.
Hence, the minimum value of the function is at
\[
  x = 1 \\
  f(1) = \dfrac{{{{(1 + 1)}^{0.6}}}}{{1 + {1^{0.6}}}} \\
   = \dfrac{{{{(2)}^{0.6}}}}{2} \\
   = {2^{(0.6 - 1)}} \\
   = {2^{ - 0.4}} \\
 \]
And the maximum value of the function will be at
\[
  x = 0 \\
  f(0) = \dfrac{{{{(1 + 0)}^{0.6}}}}{{1 + {0^{0.6}}}} \\
   = \dfrac{{{1^{0.6}}}}{1} \\
   = 1 \\
 \]
Hence, \[k = {2^{ - 0.4}}\] and \[K = 1\] so the ordered pair can be given as \[(k,K) = ({2^{ - 0.4}},1)\],

Hence, option(A) is correct answer.

Note: The maximum value of a function is the place where a function reaches its highest point, or vertex, on a graph. If your quadratic equation has a negative a term, it will also have a maximum value. If you have the graph or can draw the graph, the maximum is just the y value at the vertex of the graph.
Minimum, in mathematics, the point at which the value of a function is less than or equal to the value at any nearby point (local minimum) or at any point (absolute minimum); see extremum.