Question

Let f(x) = $\left\{ \begin{matrix} x^{2} & ifx \leq x_{0} \\ ax + b & ifx > x_{0} \end{matrix} \right.\$The values of the coefficients a and b for which the function is continuous and has a derivative at x₀, are

• A. a = x₀, b = –x₀
• B. a = 2x₀, b = –x²₀
• C. a = x²₀, b = –x₀
• D. a = x₀, b = –x²₀

Answer 1

Answer: (B)

a = 2x₀, b = –x²₀

Answer 2

For f to be continuous everywhere, we must have $x_{0}^{2}$= f(x₀)

=$\lim_{x \rightarrow {x_{0}}^{+}}$f(x) = ax₀ + b

Also, f has a derivative at x₀ if f′(x₀⁺) = f′(x₀^–)

Now, f′(x₀+) = $\lim_{h \rightarrow 0^{+}}$ $\frac{f(x_{0} + h) - f(x_{0})}{h}$

= $\lim_{h \rightarrow 0^{+}}$ $\frac{a(x_{0} + h) + b - x_{0}^{2}}{h}$

= $\lim_{h \rightarrow 0^{+}}$ $\left( \frac{ax_{0} + b - x_{0}^{2}}{h} + a \right)$= $\lim_{h \rightarrow 0^{+}}$a = a

[Q$x_{0}^{2}$= ax₀ + b]

and f′(x₀–) = $\lim_{h \rightarrow 0 -}$ $\frac{f(x_{0} + h) - f(x_{0})}{h}$

= $\lim_{h \rightarrow 0 -}$ $\frac{(x_{0} + h)^{2} - x_{0}^{2}}{h}$

= $\lim_{h \rightarrow 0 -}$ $\frac{x_{0}^{2} + h^{2} + 2x_{0}h - x_{0}^{2}}{h}$

= $\lim_{h \rightarrow 0 -}$ (h + 2x₀) = 2x₀

That is, a = 2x₀, and hence b = $x_{0}^{2}$– ax₀ = $x_{0}^{2}$–2$x_{0}^{2}$ = –$x_{0}^{2}$