Question

5 mark for review at x = 3, the function given by ( f(x) = \begin{cases} x^2 & , x < 3 \\ 6x - 9, & x ge 3 end{cases} ) is
a undefined
b continuous but not differentiable
c differentiable but not continuous
d neither continuous nor differentiable
e both continuous and differentiable

Explanation:

Step1: Check Continuity at \( x = 3 \)

First, find the left - hand limit as \( x\to3^{-} \):
For \( x\lt3 \), \( f(x)=x^{2} \). So, \( \lim_{x\to3^{-}}f(x)=\lim_{x\to3^{-}}x^{2}=3^{2} = 9 \).
Next, find the right - hand limit as \( x\to3^{+} \):
For \( x\geq3 \), \( f(x)=6x - 9 \). So, \( \lim_{x\to3^{+}}f(x)=\lim_{x\to3^{+}}(6x - 9)=6\times3-9=18 - 9 = 9 \).
Then, find the value of the function at \( x = 3 \):
Since \( x = 3 \) satisfies \( x\geq3 \), \( f(3)=6\times3-9 = 9 \).
Since \( \lim_{x\to3^{-}}f(x)=\lim_{x\to3^{+}}f(x)=f(3)=9 \), the function is continuous at \( x = 3 \).

Step2: Check Differentiability at \( x = 3 \)

First, find the left - hand derivative (derivative from the left) at \( x = 3 \):
The derivative of \( y = x^{2} \) is \( y^\prime=2x \). The left - hand derivative \( f^\prime_{-}(3)=\lim_{x\to3^{-}}\frac{f(x)-f(3)}{x - 3}=\lim_{x\to3^{-}}\frac{x^{2}-9}{x - 3} \).
We can factor \( x^{2}-9=(x - 3)(x + 3) \), so \( \lim_{x\to3^{-}}\frac{(x - 3)(x + 3)}{x - 3}=\lim_{x\to3^{-}}(x + 3)=3 + 3=6 \).
Next, find the right - hand derivative (derivative from the right) at \( x = 3 \):
The derivative of \( y = 6x-9 \) is \( y^\prime = 6 \). The right - hand derivative \( f^\prime_{+}(3)=\lim_{x\to3^{+}}\frac{f(x)-f(3)}{x - 3}=\lim_{x\to3^{+}}\frac{(6x - 9)-9}{x - 3}=\lim_{x\to3^{+}}\frac{6x-18}{x - 3}=\lim_{x\to3^{+}}\frac{6(x - 3)}{x - 3}=6 \).
Since \( f^\prime_{-}(3)=f^\prime_{+}(3) = 6 \), the function is differentiable at \( x = 3 \).

Answer:

E. both continuous and differentiable