$$ f(x)=\\begin{cases}3x + 1&\\text{for }x\\leq2\\\\5x - 3&\\text{for }x>2\\end{cases} $$ let ( f ) be the…

$$ f(x)=\\begin{cases}3x + 1&\\text{for }x\\leq2\\\\5x - 3&\\text{for }x>2\\end{cases} $$ let ( f ) be the function defined above. which of the following statements is true? a ( f ) is neither continuous nor differentiable at ( x = 2 ). b ( f ) is continuous but not differentiable at ( x = 2 ). c ( f ) is differentiable but not continuous at ( x = 2 ). d ( f ) is both continuous and differentiable at ( x = 2 )

$$ f(x)=\\begin{cases}3x + 1&\\text{for }x\\leq2\\\\5x - 3&\\text{for }x>2\\end{cases} $$ let ( f ) be the function defined above. which of the following statements is true? a ( f ) is neither continuous nor differentiable at ( x = 2 ). b ( f ) is continuous but not differentiable at ( x = 2 ). c ( f ) is differentiable but not continuous at ( x = 2 ). d ( f ) is both continuous and differentiable at ( x = 2 )

Answer

Explanation:

Step1: Check continuity at (x = 2)

  • Left - hand limit: For (x\leq2), (f(x)=3x + 1). (\lim_{x\rightarrow2^{-}}f(x)=\lim_{x\rightarrow2^{-}}(3x + 1)) Substitute (x = 2) into (3x+1): (3\times2+1=7)
  • Right - hand limit: For (x>2), (f(x)=5x - 3). (\lim_{x\rightarrow2^{+}}f(x)=\lim_{x\rightarrow2^{+}}(5x - 3)) Substitute (x = 2) into (5x - 3): (5\times2-3=7)
  • Function value: When (x = 2), (f(2)=3\times2 + 1=7) Since (\lim_{x\rightarrow2^{-}}f(x)=\lim_{x\rightarrow2^{+}}f(x)=f(2) = 7), the function (f(x)) is continuous at (x = 2)

Step2: Check differentiability at (x = 2)

  • Left - hand derivative: The derivative of (y = 3x+1) (for (x\leq2)) using the power rule ((x^n)^\prime=nx^{n - 1}) is (f^\prime(x)=3). So, (\lim_{h\rightarrow0^{-}}\frac{f(2 + h)-f(2)}{h}=\lim_{h\rightarrow0^{-}}\frac{3(2 + h)+1-7}{h}=\lim_{h\rightarrow0^{-}}\frac{6+3h + 1-7}{h}=\lim_{h\rightarrow0^{-}}\frac{3h}{h}=3)
  • Right - hand derivative: The derivative of (y = 5x-3) (for (x>2)) using the power rule is (f^\prime(x)=5). So, (\lim_{h\rightarrow0^{+}}\frac{f(2 + h)-f(2)}{h}=\lim_{h\rightarrow0^{+}}\frac{5(2 + h)-3-7}{h}=\lim_{h\rightarrow0^{+}}\frac{10+5h-3 - 7}{h}=\lim_{h\rightarrow0^{+}}\frac{5h}{h}=5)

Since (\lim_{h\rightarrow0^{-}}\frac{f(2 + h)-f(2)}{h}\neq\lim_{h\rightarrow0^{+}}\frac{f(2 + h)-f(2)}{h}), the function (f(x)) is not differentiable at (x = 2)

Answer:

B. (f) is continuous but not differentiable at (x = 2)