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For what value of the constant cc is the function ff continuous on the interval (,)(-\infty, \infty)?

f(x)={x28,if xc2x9,if x>cf(x) = \begin{cases} x^2 - 8, & \text{if } x \leq c \\ 2x - 9, & \text{if } x > c \end{cases}

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Question :

For what value of the constant cc is the function ff continuous on the interval (,)(-\infty, \infty)?

f(x)={x28,if xc2x9,if x>cf(x) = \begin{cases} x^2 - 8, & \text{if } x \leq c \\ 2x - 9, & \text{if } x > c \end{cases}

For what value of the constant c is the function f continuous on the interva | Doubtlet.com

Solution:

Neetesh Kumar

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

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

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

For xcx \leq c, the function is given by f(x)=x28f(x) = x^2 - 8. So, the left-hand limit as xcx \to c^- is:

limxcf(x)=c28\lim_{x \to c^-} f(x) = c^2 - 8

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

For x>cx > c, the function is given by f(x)=2x9f(x) = 2x - 9. So, the right-hand limit as xc+x \to c^+ is:

limxc+f(x)=2c9\lim_{x \to c^+} f(x) = 2c - 9

Step 3: Set the limits equal to each other

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

c28=2c9c^2 - 8 = 2c - 9

Step 4: Solve for cc

Simplify the equation:

c28=2c9c^2 - 8 = 2c - 9

Rearrange the terms:

c22c+1=0c^2 - 2c + 1 = 0

This is a quadratic equation. We can factor it:

(c1)2=0(c - 1)^2 = 0

Thus, c=1c = 1.


Therefore, the value of cc that makes the function continuous is:

c=1c =\boxed{1}



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