Mathematics: which of the following functions is continuous at x = 5? (5 points) f of x equals the quantity x squared minus twenty five divided by the quantity x plus five f of x equals the quotient of the quantity x squared minus 25 and the quantity x minus 5 for x not equal to 5 and equals 20 for x equals 5 f of x equals the quotient of the quantity x squared minus 25 and the quantity x minus 5 for x not equal to 5 and equals 0 for x equals 5 all are continuous at x = 5

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which of the following functions is continuous

freddyfriendo2364

12.01.2022

$f(x)=\frac{x^2-25}{x+5}$

Step-by-step explanation:

Given:

We need to check the continuity of the functions at x=5 .

Option 1:

$f(x)=\frac{x^2-25}{x+5}$

Plug in 5 for 'x' and check the value of f(x) .

$f(5)=\frac{5^2-25}{5+5}=\frac{25-25}{10}=0$

$\lim_{x \to 5} \frac{x^2-25}{x+5}= \lim_{x \to 5} \frac{(x-5)(x+5)}{x+5}= \lim_{x \to 5} (x-5)=5-5=0$

So, the function is defined at and around x=5 and equal to 0. So, it is continuous at x=5 .

Option 2:

$f(x)=\left \{ {{\frac{x^2-25}{x-5}}\ x\ne 5 \atop {20}\ \ \ x=5} \right$

The given function is defined at x=5 and is equal to 20 but it has a different value around 5.

$\lim_{x \to 5} \frac{x^2-25}{x-5}= \lim_{x \to 5} \frac{(x-5)(x+5)}{x-5}= \lim_{x \to 5} (x+5)=5+5=10$

So, the function has a value of 10 around x=5 and equal to 20 at x=5 . So, it is not continuous at x=5 .

Option 3:

$f(x)=\left \{ {{\frac{x^2-25}{x-5}}\ x\ne 5 \atop {0}\ \ \ x=5} \right$

The given function is defined at x=5 and is equal to 0 but it has a different value around 5.

$\lim_{x \to 5} \frac{x^2-25}{x-5}= \lim_{x \to 5} \frac{(x-5)(x+5)}{x-5}= \lim_{x \to 5} (x+5)=5+5=10$

So, the function has a value of 10 around x=5 and equal to 0 at x=5 . So, it is not continuous at x=5 .

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