Question

If f(x) = $\sqrt{x + 3 - 4\sqrt{x - 1}} + \sqrt{x + 8 - 6\sqrt{x - 1}}$ then f ′(x) at

x = 1.5 is

• A. 0
• B. –$\sqrt{2}$
• C. –$\sqrt{3}$
• D. – 4

Answer 1

Answer: (B)

–$\sqrt{2}$

Answer 2

f(x) =$\sqrt{(x - 1) + 4 - 4\sqrt{x - 1}} + \sqrt{(x - 1) + 9 - 6\sqrt{x - 1}}$

=$\left| \sqrt{x - 1} - 2 \right| + \left| \sqrt{x - 1} - 3 \right|$

at x = 1.5

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

= 5 – 2 $\sqrt{x - 1}$

f ′(x) = $\frac{- 2}{2\sqrt{x - 1}}$

∴ f ′(1.5) = $\frac{- 1}{\sqrt{\frac{3}{2} - 1}} = - \sqrt{2}$