Question

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

Answer 1

c

Explanation:

edge 2022

Answer 2

`f(x)=(x^2-25)/(x+5)`

Explanation:

Given:

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

Option 1:

`f(x)=(x^2-25)/(x+5)`

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

`f(5)=(5^2-25)/(5+5)=(25-25)/(10)=0`

`\lim_(x \to 5) (x^2-25)/(x+5)= \lim_(x \to 5) ((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 \{ {{(x^2-25)/(x-5)}\ x\\e 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) (x^2-25)/(x-5)= \lim_(x \to 5) ((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 \{ {{(x^2-25)/(x-5)}\ x\\e 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) (x^2-25)/(x-5)= \lim_(x \to 5) ((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`.