The function defined by f x - Veft matrix kx -1- -if - X leq 5113 x-5-8 - i f - x-5A- 5 - 9B- 9 - 5C- 4 - 9D- 9 - 4

The correct option is B 9/5 LHL = RHL ⇒ 5K + 1 = 10 ⇒ k = 9/5 ...

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Continuity at a Boundary

The function ...

Question

The function defined by $f (x) = {\begin{matrix} k x + 1, & i f x \leq 5 3 x - 5, & i f x > 5 \end{matrix}$ is continuous at x = 5, the k =

5/9

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9/5

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4/9

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9/4

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Solution

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Similar questions

f (x) = \frac{2 x + 1}{x - 9}

f (x)

f (x) = \frac{x^{4} - 64 x}{\sqrt{x^{2} + 9} - 5}

x \neq 4

= 3

x = 4

f (x)

x = 4

\int_{1}^{4} f (x) d x, where f (x) = \{\begin{cases} 4 x + 3, if 1 \leq x \leq 2 \\ 3 x + 5, if 2 \leq x \leq 4 \end{cases}

\int_{0}^{9} f (x) d x, where f (x) \{\begin{matrix} \sin x & , & 0 \leq x \leq π / 2 \\ 1 & , & π / 2 \leq x \leq 3 \\ e^{x - 3} & , & 3 \leq x \leq 9 \end{matrix}

\int_{1}^{4} f (x) d x, where f (x) = \{\begin{matrix} 7 x + 3 & , & if 1 \leq x \leq 3 \\ 8 x & , & if 3 \leq x \leq 4 \end{matrix}

The function is defined by f(x) = ${\begin{matrix} K x = 1, i f x \leq 5 3 x - 5, i f x > 5 \end{matrix}$ at x = 5.

f (x) = |x^{2} - 9 x + 20|

- 2 x + 9 for all x \in R

2 x - 9 if 4 < x < 5

- 2 x + 9, if 4 < x < 5

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