Question

$$ f(x)=\
LATEXBLOCK0
$$

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 )

Explanation:

Step1: Check continuity at \(x = 2\)

• Left - hand limit:

For \(x\leq2\), \(f(x)=3x + 1\). \(\lim_{x
ightarrow2^{-}}f(x)=\lim_{x
ightarrow2^{-}}(3x + 1)\)
Substitute \(x = 2\) into \(3x+1\): \(3\times2+1=7\)

• Right - hand limit:

For \(x>2\), \(f(x)=5x - 3\). \(\lim_{x
ightarrow2^{+}}f(x)=\lim_{x
ightarrow2^{+}}(5x - 3)\)
Substitute \(x = 2\) into \(5x - 3\): \(5\times2-3=7\)

• Function value:

When \(x = 2\), \(f(2)=3\times2 + 1=7\)
Since \(\lim_{x
ightarrow2^{-}}f(x)=\lim_{x
ightarrow2^{+}}f(x)=f(2) = 7\), 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, \(\lim_{h
ightarrow0^{-}}\frac{f(2 + h)-f(2)}{h}=\lim_{h
ightarrow0^{-}}\frac{3(2 + h)+1-7}{h}=\lim_{h
ightarrow0^{-}}\frac{6+3h + 1-7}{h}=\lim_{h
ightarrow0^{-}}\frac{3h}{h}=3\)

• Right - hand derivative:

The derivative of \(y = 5x-3\) (for \(x>2\)) using the power rule is \(f^\prime(x)=5\). So, \(\lim_{h
ightarrow0^{+}}\frac{f(2 + h)-f(2)}{h}=\lim_{h
ightarrow0^{+}}\frac{5(2 + h)-3-7}{h}=\lim_{h
ightarrow0^{+}}\frac{10+5h-3 - 7}{h}=\lim_{h
ightarrow0^{+}}\frac{5h}{h}=5\)

Since \(\lim_{h
ightarrow0^{-}}\frac{f(2 + h)-f(2)}{h}
eq\lim_{h
ightarrow0^{+}}\frac{f(2 + h)-f(2)}{h}\), the function \(f(x)\) is not differentiable at \(x = 2\)

Answer:

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