Question

Find numbers a and b such that the given function g is differentiable at 1.

\(g\left( x \right) = \left\{ {\begin{aligned}{a{x^{\bf{3}}} - {\bf{3}}x}&{{\bf{if}}\,\,x \le {\bf{1}}}\\{b{x^{\bf{2}}} + {\bf{2}}}&{{\bf{if}}\,\,x > {\bf{1}}}\end{aligned}} \right.\)

Short answer

The required values are \(a = - 7\) and \(b = - 12\).

Step-by-step solution

Find the derivative of g for \(x \le {\bf{1}}\)

Differentiate the function \(g\left( x \right) = a{x^3} - 3x\)\(\left( {x \le 1} \right)\).

\(\begin{aligned}g'\left( x \right) &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {a{x^3} - 3x} \right)\\ &= 3a{x^2} - 3\end{aligned}\)

For \(x = {1^ - }\),

\(\begin{aligned}g'\left( {{1^ - }} \right) &= 3a{\left( 1 \right)^2} - 3\\ &= 3a - 3\end{aligned}\)

Find the derivative of g for \(x > {\bf{1}}\)

Differentiate the function \(g\left( x \right) = b{x^2} + 2\)\(\left( {x > 1} \right)\).

\(\begin{aligned}g'\left( x \right) &= \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {b{x^2} + 2} \right)\\ &= 2bx\end{aligned}\)

For \(x = {1^ + }\),

\(\begin{aligned}g'\left( {{1^ + }} \right) &= 2b\left( 1 \right)\\ &= 2b\end{aligned}\)

As \(g\left( x \right)\) is differentiableat \(x = 1\), then

\(\begin{aligned}g'\left( {{1^ - }} \right) &= g'\left( {{1^ + }} \right)\\3a - 3 &= 2b\end{aligned}\)

Find relation between a and b using continuity of g

Find the value of \(g\left( {{1^ - }} \right)\).

\(\begin{aligned}g\left( {{1^ - }} \right) &= a{\left( 1 \right)^3} - 3\left( 1 \right)\\ &= a - 3\end{aligned}\)

Find the value of \(g\left( {{1^ + }} \right)\).

\(\begin{aligned}g\left( {{1^ + }} \right) &= b{\left( 1 \right)^2} + 2\\ &= b + 2\end{aligned}\)

As the function is continuous at \(x = 1\), therefore

\(\begin{aligned}a - 3 &= b + 2\\a &= b + 5\end{aligned}\)

Find the values of a and b

Solve the equations \(3a - 3 = 2b\) and \(a = b + 5\).

\(\begin{aligned}3\left( {b + 5} \right) - 3 &= 2b\\3b + 15 - 3 &= 2b\\b &= - 12\end{aligned}\)

And

\(\begin{aligned}a &= - 12 + 5\\ &= - 7\end{aligned}\)

Thus, the values are \(a = - 7\) and \(b = - 12\).