Question

A function g, defined for all positive real numbers, satisfies the following two conditions g(1) = 1 and g´(x²) = x³ for all

x > 0, then g(4) is equal to

• A. $\frac{61}{5}$
• B. $\frac{64}{5}$
• C. $\frac{67}{5}$
• D. 14

Answer 1

Answer: (C)

$\frac{67}{5}$

Answer 2

Let x² = t; g¢ (t) = t^3/2; g(t) = $\frac{2}{5}$t^5/2 + C

g(1) = 1 = $\frac{2}{5}$ + C ̃ C = 1 – $\frac{2}{5}$ = $\frac{3}{5}$

so g(t) = $\frac{2}{5}$t^5/2 + $\frac{3}{5}$

g(4) = $\frac{2}{5}$.4^5/2 + $\frac{3}{5}$ = $\frac{64 + 3}{5}$ = $\frac{67}{5}$