Question

Examine the function f(x) = \( \begin{cases} |x - 3|; &x \geq1\\ \frac{x^2}{4} - \frac{3x}{2} + \frac{13}{4}; & x < 1 \end{cases} \) for continuity at x = 1 and 3.

Answer 1

For continuity at x = 1

= | 1 + 0 – 3 |

= | – 2 |

= 2

Value of function f(x) at x = 1,

f(1) = | 1 – 3 |

= | -2 |

= 2

∴ f (1 – 0) = f(1 + 0) = f (1) = 2

So, f(x) is continue at f (x), x = 1

For continuity at x = 3

Left hand limit

f (3 – 0) = lim_h→0 f (3 – h)

= lim_h→0 – {(3-A)-3}

= lim_h→0 (3 – 3 + h)

= 0

Right hand limit

f(3 + 0) = lim_h→0 f(3 + h)

= lim_h→0 (3 + h – 3)
= 0

f(3) = 3 – 3 = 0

∴ f (3 – 0) = f(3 + 0)

= f(0) = o

Hence, function is continue at x = 3.