For what value of the constant $c$ is the function $f$ continuous on the interval $(-\infty, \infty)$?

$f(x) = \begin{cases} x^2 - 8, & \text{if } x \leq c \\ 2x - 9, & \text{if } x > c \end{cases}$

Neetesh Kumar | October 15, 2024

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Assignment Question on Continuous Function
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Step-by-step solution:

To ensure that $f(x)$ is continuous at $x = c$, the function values from both sides of $c$ must be equal at $x = c$. Specifically, we need:

$\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)$

Step 1: Evaluate the left-hand limit as $x \to c^-$

For $x \leq c$, the function is given by $f(x) = x^2 - 8$. So, the left-hand limit as $x \to c^-$ is:

$\lim_{x \to c^-} f(x) = c^2 - 8$

Step 2: Evaluate the right-hand limit as $x \to c^+$

For $x > c$, the function is given by $f(x) = 2x - 9$. So, the right-hand limit as $x \to c^+$ is:

$\lim_{x \to c^+} f(x) = 2c - 9$

Step 3: Set the limits equal to each other

To ensure continuity at $x = c$, we set the left-hand and right-hand limits equal to each other:

$c^2 - 8 = 2c - 9$

Step 4: Solve for $c$

Simplify the equation:

$c^2 - 8 = 2c - 9$

Rearrange the terms:

$c^2 - 2c + 1 = 0$

This is a quadratic equation. We can factor it:

$(c - 1)^2 = 0$

Thus, $c = 1$.

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Therefore, the value of $c$ that makes the function continuous is:

$c =\boxed{1}$

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