# Question #54b57

Jan 24, 2018

$\left(\frac{5}{12} , \frac{13}{12}\right)$

#### Explanation:

$f \left(x\right) = \sqrt{3 x + 1} - x$

$f \left(x\right) = {\left(3 x + 1\right)}^{\frac{1}{2}} - x$

Take the first derivative:

$f ' \left(x\right) = \left(\frac{1}{2}\right) {\left(3 x + 1\right)}^{- \frac{1}{2}} \cdot 3 - 1$

$f ' \left(x\right) = \frac{3}{2 \sqrt{3 x + 1}} - 1$

Set the first derivative to zero to solve for critical point(s):

$\frac{3}{2 \sqrt{3 x + 1}} - 1 = 0$

$\frac{3 - 2 \sqrt{3 x + 1}}{2 \sqrt{3 x + 1}} = 0$ => denominator can't be zero:

$3 = 2 \sqrt{3 x + 1}$ => square both sides:

$9 = 4 \left(3 x + 1\right)$

$9 = 12 x + 4$

$5 = 12 x$

$x = \frac{5}{12}$ => find the y value at this point:

$f \left(\frac{5}{12}\right) = \sqrt{\frac{5}{4} + 1} - \frac{5}{12} = \sqrt{\frac{9}{4}} - \frac{5}{12}$

$= \frac{3}{2} - \frac{5}{12} = \frac{18}{12} - \frac{5}{12} = \frac{13}{12}$

Thus our critical point is: $\left(\frac{5}{12} , \frac{13}{12}\right)$ now to determine the

nature of this point we use the 2nd derivative test:

$f ' ' \left(x\right) = - \frac{9}{4 {\left(3 x + 1\right)}^{\frac{3}{2}}}$

$f ' ' \left(\frac{5}{12}\right) = - \frac{2}{3} < 0$

This proves that our critical point is indeed a global/absolute

maximum.