If f x = e 2 x 2 + x :   x > 0 a x + b :   x ≤ 0 is differentiable at x = 0 , then

Step-by-step solution

f ′ 0 − = lim h → 0 + f 0 − h − f 0 − h ⇒ lim h → 0 + f − a h + b − b − h = a f ' 0 + = lim h → 0 + f 0 + h − f 0 h ⇒ lim h → 0 + e 2 h 2 + h − b h     . . .   1 For limit to exists, it must be 0 0 form ∴ e 2 0 + 0 - b = 0 ⇒ b = 1 ... 2 From 1 ,   f ' 0 + = lim h → 0 + e 2 h 2 + h − 1 h = lim h → 0 + e 2 h 2 + h − 1 2 h 2 + h × 2 h 2 + h h = 1 × 1 = 1 ...