Question

1. $f(x)=\begin{cases}kx^{2}&xleq3\\4x - 11&x>3end{cases}$

Explanation:

1. Explanation:

• For a function \(y = f(x)\) to be continuous at \(x = a\), the left - hand limit \(\lim_{x

ightarrow a^{-}}f(x)\) must be equal to the right - hand limit \(\lim_{x
ightarrow a^{+}}f(x)\) and also equal to \(f(a)\). Here, \(a = 3\), and we want to find \(k\) such that \(f(x)\) is continuous at \(x = 3\).

• Step 1: Calculate the left - hand limit
• The left - hand limit as \(x

ightarrow3^{-}\) (approaching 3 from the left side, where \(x\leq3\)) is \(\lim_{x
ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{-}}kx^{2}\).

• Substitute \(x = 3\) into \(kx^{2}\), we get \(\lim_{x

ightarrow3^{-}}kx^{2}=k\times3^{2}=9k\).

• Step 2: Calculate the right - hand limit
• The right - hand limit as \(x

ightarrow3^{+}\) (approaching 3 from the right side, where \(x > 3\)) is \(\lim_{x
ightarrow3^{+}}f(x)=\lim_{x
ightarrow3^{+}}(4x - 11)\).

• Substitute \(x = 3\) into \(4x-11\), we get \(\lim_{x

ightarrow3^{+}}(4x - 11)=4\times3-11=12 - 11 = 1\).

• Step 3: Set the left - hand limit equal to the right - hand limit
• Since the function is continuous at \(x = 3\), we set \(\lim_{x

ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{+}}f(x)\).

• So, \(9k=1\).
• Solve for \(k\) by dividing both sides of the equation by 9: \(k=\frac{1}{9}\).

1. Answer:

\(k=\frac{1}{9}\)

Answer:

1. Explanation:

• For a function \(y = f(x)\) to be continuous at \(x = a\), the left - hand limit \(\lim_{x

ightarrow a^{-}}f(x)\) must be equal to the right - hand limit \(\lim_{x
ightarrow a^{+}}f(x)\) and also equal to \(f(a)\). Here, \(a = 3\), and we want to find \(k\) such that \(f(x)\) is continuous at \(x = 3\).

• Step 1: Calculate the left - hand limit
• The left - hand limit as \(x

ightarrow3^{-}\) (approaching 3 from the left side, where \(x\leq3\)) is \(\lim_{x
ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{-}}kx^{2}\).

• Substitute \(x = 3\) into \(kx^{2}\), we get \(\lim_{x

ightarrow3^{-}}kx^{2}=k\times3^{2}=9k\).

• Step 2: Calculate the right - hand limit
• The right - hand limit as \(x

ightarrow3^{+}\) (approaching 3 from the right side, where \(x > 3\)) is \(\lim_{x
ightarrow3^{+}}f(x)=\lim_{x
ightarrow3^{+}}(4x - 11)\).

• Substitute \(x = 3\) into \(4x-11\), we get \(\lim_{x

ightarrow3^{+}}(4x - 11)=4\times3-11=12 - 11 = 1\).

• Step 3: Set the left - hand limit equal to the right - hand limit
• Since the function is continuous at \(x = 3\), we set \(\lim_{x

ightarrow3^{-}}f(x)=\lim_{x
ightarrow3^{+}}f(x)\).

• So, \(9k=1\).
• Solve for \(k\) by dividing both sides of the equation by 9: \(k=\frac{1}{9}\).

1. Answer:

\(k=\frac{1}{9}\)