Curve y = $\sqrt{3\mathit{x}-2}$
$\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ = $\frac{1}{2}{\left(3\mathit{x}-2\right)}^{-\frac{1}{2}}\times 3$
⇒ $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ = $\frac{3}{2\sqrt{\left(3\mathit{x}-2\right)}}$...(1)
Since, the tangent is parallel to the line $4\mathit{x}-2\mathit{y}$ = - 5
Therefore, slope of tangent can be obtained from equation
y = $\frac{4\mathit{x}}{2}+\frac{5}{2}$
Slope = 2
⇒ $\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}$ = 2
Comparing equations (1) and (2), we have,
$\frac{3}{2}\times \frac{1}{\sqrt{\left(3\mathit{x}-2\right)}}$ = 2
$\frac{1}{\sqrt{\left(3\mathit{x}-2\right)}}$ = $\frac{4}{3}$
⇒ $\frac{1}{3\mathit{x}-2}$ = $\frac{16}{9}$
⇒ 9 = 48x - 32
⇒ x = $\frac{41}{48}$
We have y = $\sqrt{\left(3\mathit{x}-2\right)}$
Thus, substituting the value of x in the above equation,
y = $\sqrt{3\times \frac{41}{48}-2}$
⇒ y = $\sqrt{\frac{41}{16}-2}$
⇒ y = $\sqrt{\frac{41-32}{16}}$
⇒ y = $\sqrt{\frac{9}{16}}$
⇒ y = $\frac{3}{4}$
Equation of tangent is
$\left(\mathit{y}-\frac{3}{4}\right)$ = 2$\left(\mathit{x}-\frac{41}{48}\right)$
⇒ $\left(\mathit{y}-\frac{3}{4}\right)$ = $2\mathit{x}-\frac{41}{24}$
⇒ y = $2\mathit{x}-\frac{41}{24}+\frac{3}{4}$
⇒ y = $2\mathit{x}-\frac{41}{24}+\frac{18}{24}$
⇒ y = $2\mathit{x}-\frac{23}{24}$
⇒ $24\mathit{y}=48\mathit{x}-23$
⇒ $48\mathit{x}-24\mathit{y}-23+0$
OR
$\mathit{f}\left(\mathit{x}\right)$ = ${\mathit{x}}^3+\frac{1}{{\mathit{x}}^3},\mathit{x}\ne 0$
⇒ $\mathit{f}\prime \left(\mathit{x}\right)=3{\mathit{x}}^2-3{\mathit{x}}^{-4}=3\left({\mathit{x}}^2-\frac{1}{{\mathit{x}}^4}\right)$
⇒ $\mathit{f}\prime \left(\mathit{x}\right)=3{\mathit{x}}^2-3{\mathit{x}}^4=\frac{3}{{\mathit{x}}^4}\left({\mathit{x}}^6-1\right)$
⇒ $\mathit{f}\prime \left(\mathit{x}\right)=\frac{3}{{\mathit{x}}^4}\left({\mathit{x}}^2-1\right)\left({\mathit{x}}^4+{\mathit{x}}^2+1\right)$
⇒ f'(x) = 3 $\left(\frac{{\mathit{x}}^4+{\mathit{x}}^2+1}{{\mathit{x}}^4}\right)\left({\mathit{x}}^2-1\right)$
(i) For an increasing function, we should have,
f ' (x) > 0
⇒ 3 $\left(\frac{{\mathit{x}}^4+{\mathit{x}}^2+1}{{\mathit{x}}^4}\right)\left({\mathit{x}}^2-1\right)$ > 0
⇒ $\left({\mathit{x}}^2-1\right)$ > 0 [Since 3 $\left(\frac{{\mathit{x}}^4+{\mathit{x}}^2+1}{{\mathit{x}}^4}\right)$ > 0]
⇒ (x - 1) (x + 1) > 0
⇒ x ∊ (- ∞ , - 1) ∪ x ∊ (1 , ∞)
So, f(x) is increasing on (- ∞ , - 1) ∪ (1 , ∞)
(ii) For a decreasing function, we should have f’(x) < 0
f ' (x) < 0
⇒ 3 $\left(\frac{{\mathit{x}}^4+{\mathit{x}}^2+1}{{\mathit{x}}^4}\right)\left({\mathit{x}}^2-1\right)$ < 0
⇒ $\left({\mathit{x}}^2-1\right)$ < 0 [Since 3 $\left(\frac{{\mathit{x}}^4+{\mathit{x}}^2+1}{{\mathit{x}}^4}\right)\left({\mathit{x}}^2-1\right)$ > 0]
⇒ (x - 1) (x + 1) < 0
⇒ ∊ (- 1 , 0) ∪ x ∊ (0 , 1)
So f(x) is decreasing on (- 1 , 0) ∪ (0 , 1)