Chapter 14: Q. 7 (page 907)

Determine the value for k that will make the function continuous at $c = 4$ $f (x) = \{\begin{cases} \frac{x^{2} - 9}{x + 3} i f x \leq 4 \\ k x + 5 i f x > 4 \end{cases}$

Short Answer

Expert verified

The value of $k =$ $- 1$

Step by step solution

Step 1. Value of the function at c=4

The function where $c = 4$ is

$f (x) = \frac{x^{2} - 9}{x + 3}$

Substituting the values,

$f (4) = \frac{4^{2} - 9}{4 + 3} = \frac{7}{7} = 1$

Step 2. Apply the limit for c<4

On applying the limit, we get,

$\lim_{x \to 4^{-}} \frac{x^{2} - 9}{x + 3} = \lim_{x \to 4^{-}} (\frac{(x + 3) (x - 3)}{x + 3})$

$= \lim_{x \to 4^{-}} (x - 3) = 4 - 3 = 1$

Step 3. Apply the limit for c>4

On applying the limit,

$\lim_{x \to 4^{+}} (k x + 5) = \lim_{x \to 4^{+}} k x + \lim_{x \to 4^{+}} 5 = k \lim_{x \to 4^{+}} x + \lim_{x \to 4^{+}} 5 = 4 k + 5$

Step 4. Value of k

We know, if a continuous limit exists then

$\lim_{x \to 4^{-}} f (x) = \lim_{x \to 4^{+}} f (x) = \lim_{x \to 4} f (x) 1 = 4 k + 5 - 4 = 4 k - 1 = k$

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Determine the value for k that will make the function continuous at $c = 4$ $f (x) = \{\begin{cases} \frac{x^{2} - 9}{x + 3} i f x \leq 4 \\ k x + 5 i f x > 4 \end{cases}$

Short Answer

Step by step solution

Step 1. Value of the function at c=4

Step 2. Apply the limit for c<4

Step 3. Apply the limit for c>4

Step 4. Value of k

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Determine the value for k that will make the function continuous atc=4 f(x)=x2-9x+3ifx≤4kx+5ifx>4

Short Answer

Step by step solution

Step 1. Value of the function at c=4

Step 2. Apply the limit for c<4

Step 3. Apply the limit for c>4

Step 4. Value of k

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Mechanics Maths

Calculus

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Study anywhere. Anytime. Across all devices.

Determine the value for k that will make the function continuous at $c = 4$ $f (x) = \{\begin{cases} \frac{x^{2} - 9}{x + 3} i f x \leq 4 \\ k x + 5 i f x > 4 \end{cases}$