What are the local extrema of f(x)= 4x^2-2x+x/(x-1/4)?

Apr 18, 2017

${f}_{\min} = f \left(\frac{1}{4} + {2}^{- \frac{5}{3}}\right) = \frac{{2}^{\frac{2}{3}} + 3 + {2}^{\frac{5}{3}}}{4.}$

Explanation:

Observe that, f(x)=4x^2-2x+x/(x-1/4); x in RR-{1/4}.

=4x^2-2x+1/4-1/4+{(x-1/4)+1/4}/(x-1/4); xne1/4

=(2x-1/2)^2-1/4+{(x-1/4)/(x-1/4)+(1/4)/(x-1/4)}; xne1/4

=4(x-1/4)^2-1/4+{1+(1/4)/(x-1/4)}; xne1/4

 :. f(x)=4(x-1/4)^2+3/4+(1/4)/(x-1/4); xne1/4.

Now, for Local Extrema, $f ' \left(x\right) = 0 ,$ and,

$f ' ' \left(x\right) > \mathmr{and} < 0 , \text{ according as "f_(min) or f_(max)," resp.}$

$f ' \left(x\right) = 0$

$\Rightarrow 4 \left\{2 \left(x - \frac{1}{4}\right)\right\} + 0 + \frac{1}{4} \left\{\frac{- 1}{x - \frac{1}{4}} ^ 2\right\} = 0. . . \left(\ast\right)$

$\Rightarrow 8 \left(x - \frac{1}{4}\right) = \frac{1}{4 {\left(x - \frac{1}{4}\right)}^{2}} , \mathmr{and} , {\left(x - \frac{1}{4}\right)}^{3} = \frac{1}{32} = {2}^{-} 5.$

$\Rightarrow x = \frac{1}{4} + {2}^{- \frac{5}{3}}$

Further, $\left(\ast\right) \Rightarrow f ' ' \left(x\right) = 8 - \frac{1}{4} \left\{- 2 {\left(x - \frac{1}{4}\right)}^{-} 3\right\} , \text{ so that, }$

$f ' ' \left(\frac{1}{4} + {2}^{- \frac{5}{3}}\right) = 8 + \left(\frac{1}{2}\right) {\left({2}^{- \frac{5}{3}}\right)}^{-} 3 > 0$

$\text{Therefore, } {f}_{\min} = f \left(\frac{1}{4} + {2}^{- \frac{5}{3}}\right)$

$= 4 {\left({2}^{- \frac{5}{3}}\right)}^{2} + \frac{3}{4} + \frac{\frac{1}{4}}{{2}^{- \frac{5}{3}}} = {2}^{2} \cdot {2}^{- \frac{10}{3}} + \frac{3}{4} + {2}^{- 2} \cdot {2}^{\frac{5}{3}}$

$= \frac{1}{2} ^ \left(\frac{4}{3}\right) + \frac{3}{2} ^ 2 + \frac{1}{2} ^ \left(\frac{1}{3}\right) = \frac{{2}^{\frac{2}{3}} + 3 + {2}^{\frac{5}{3}}}{4.}$

Thus, ${f}_{\min} = f \left(\frac{1}{4} + {2}^{- \frac{5}{3}}\right) = \frac{{2}^{\frac{2}{3}} + 3 + {2}^{\frac{5}{3}}}{4.}$

Enjoy Maths.!