Question

Find at what value the function \[f(x) = \dfrac{2}{x} + \dfrac{x}{2}\] has a local minimum.
A.\[x = - 2\]
B.\[x = 0\]
C.\[x = 1\]
D.\[x = 2\]

Answer 1

Hint: First differentiate the given equation with respect to x, then equate the derived equation \[\dfrac{1}{2} - \dfrac{2}{{{x^2}}}\] to zero to obtain the critical points. Differentiate the function \[\dfrac{1}{2} - \dfrac{2}{{{x^2}}}\] , then substitute 2 for x in the equation \[f''(x) = \dfrac{4}{{{x^3}}}\]and observe whether the function is less than zero or greater than zero.

Formula used:
\[\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}\] .
A function f(x) at any point ‘a’ has a minimum value at ‘a’ if \[f''(a) > 0\] .

Complete step by step solution:
Differentiate \[f(x) = \dfrac{2}{x} + \dfrac{x}{2}\] with respect to x.
\[\begin{array}{I}f'(x) = \dfrac{d}{{dx}}\left[ {2{x^{ - 1}} + \dfrac{1}{2}x} \right]\\ = - 2{x^{ - 2}} + \dfrac{1}{2}\\ = \dfrac{1}{2} - \dfrac{2}{{{x^2}}}\end{array}\]
Now, equate \[f'(x)\] to zero, to obtain the critical points.
\[\dfrac{1}{2} - \dfrac{2}{{{x^2}}} = 0\]
\[\dfrac{2}{{{x^2}}} = \dfrac{1}{2}\]
\[{x^2} = 4\]
\[x = \pm 2\]
Differentiate \[f'(x)\] once again.
\[f''(x) = \dfrac{d}{{dx}}\left[ {\dfrac{1}{2} - \dfrac{2}{{{x^2}}}} \right]\]
\[f''(x) = 0 - 2\dfrac{{\left( { - 2} \right)}}{{{x^3}}}\]
\[f''(x) = \dfrac{4}{{{x^3}}}\]
Now,
\[f''(2) = \dfrac{4}{{{2^3}}}\]
\[ = \dfrac{4}{8}\]
\[ = \dfrac{1}{2} > 0\]
And
\[f''(-2) = \dfrac{4}{{{{\left( { - 2} \right)}^3}}}\]
\[ = - \dfrac{4}{8}\]
\[ = - \dfrac{1}{2} < 0\]

Therefore, f(x) has a minima at x=2.

The correct option is D.

Note Sometimes students get confused that if \[f''(2) < 0\] then f(x) is minimum or if \[f''(2) > 0\]then f(x) is minimum? So, just remember that “a function f(x) at any point ‘a’ has a minimum value at ‘a’ if \[f''(a) > 0\] “.