If f(x) = \ cases e^ 2x^ 2 + x & : x > 0 ax + b & : x ≤ 0 cases )...

f^{′} \left({0}^{−}\right) = \lim_{h \to {0}^{+}} \dfrac{f\left(0 - h\right) - f\left(0\right)}{- h} ⇒ \lim_{h \to {0}^{+}} \dfrac{f\left(- ah + b\right) - \left(b\right)}{- h} = a$<br /><br />$f^{'}\left({0}^{+}\right) = \lim_{h \to {0}^{+}}\dfrac{f\left(0 + h\right) - f \left(0\right)}{h} ⇒ \lim_{h \to {0}^{+}}\dfrac{ e^{2h^{2} + h} - b}{h} . . . \left(1\right)$<br /><br />For limit to exists, it must be$\dfrac{0}{0}form

$∴ e^{2 \left(0\right) + 0}$ - b = 0 ⇒ b = 1... \left(2\right)<br /><br />From \left(1\right), f^{'}\left({0}^{+}\right) = \lim_{h \to {0}^{+}}\dfrac{ e^{2h^{2} + h} - 1}{h}

$= \lim_{h \to {0}^{+}}e^{2h^{2} + h} - 1 \left(2h^{2} + h\right)\times\left(\dfrac{2h^{2} + h}{h}\right)$ =1 \times1 = 1$... \left(3\right)<br /><br />$∵ f \left( x\right)$is differentiable at$x = 0$<br /><br />$∴ f^{'} \left({0}^{−}\right)$= f^{'} \left({0}^{+}\right)$

$⇒ a = 1$