Discussion
A function \(f\) is defined on R by :
\[ f(x) = \begin{cases} x^{2} & \text{if } x \le 3 \\ ax + b & \text{if } x \gt 3 \end{cases} \]
Then, for which values of \(a\) and \(b\), \(f'(3)\) exists ?
Answer: A
The derivative at 3 exists for \(a = 6,\ b = -9\).
Two conditions are needed. First, \(f\) must be continuous at 3, so the right-hand value \(3a + b\) must equal \(3^{2} = 9\). Second, the one-sided derivatives must agree: on the left \(f'(x) = 2x\) gives 6, and on the right the slope is \(a\).
\[ \begin{aligned} 3a + b &= 9 \\ a &= 6 \\ b &= 9 - 18 = -9 \end{aligned} \]
So the line \(6x - 9\) is the tangent to \(x^{2}\) at 3. The pair \(a = 0, b = 9\) is continuous but has slope 0 on the right; \(a = 3, b = 0\) is also continuous (\(3a + b = 9\)) but has slope 3, not 6; and \(a = 18, b = -9\) gives \(3a + b = 45\), so \(f\) jumps at 3.
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