# If f(x) is continuous and differentiable and f(x) = ax^4 + 5x; x<=2 and bx^2 - 3x; x> 2, then how do you find b?

Nov 7, 2016

$b = \frac{15}{2}$

#### Explanation:

As $f \left(x\right)$ is continuous at $x = 2$, we have

${\lim}_{x \to {2}^{-}} f \left(x\right) = {\lim}_{x \to {2}^{+}} f \left(x\right)$

$\implies a \left({2}^{4}\right) + 5 \left(2\right) = b \left({2}^{2}\right) - 3 \left(2\right)$

$\implies 16 a + 10 = 4 b - 6$

$\implies a = \frac{1}{4} b - 1$

As $f \left(x\right)$ is differentiable at $x = 2$, the limit $f ' \left(2\right) = {\lim}_{x \to 2} \frac{f \left(x\right) - f \left(2\right)}{x - 2}$ must exist. We can tell what the one sided limits will evaluate to by calculating the derivatives of the components of the piecewise defined functions on either side of $2$.

${\lim}_{x \to {2}^{-}} \frac{f \left(x\right) - f \left(2\right)}{x - 2} = {\lim}_{x \to {2}^{+}} \frac{f \left(x\right) - f \left(2\right)}{x - 2}$

$\implies 4 a \left({2}^{3}\right) + 5 = 2 b \left(2\right) - 3$

$\implies 32 a + 5 = 4 b - 3$

Substituting in $a = \frac{1}{4} b - 1$, we have

$32 \left(\frac{1}{4} b - 1\right) + 5 = 4 b - 3$

$\implies 8 b - 27 = 4 b - 3$

$\implies 4 b = 30$

$\therefore b = \frac{15}{2}$