Question

Solve the problem. To what new value should f(1) be changed to remove the discontinuity? f(x)=⎩⎨⎧​x2+21x+2​x<1x=2x>1​ 4 2 1 3

Answer 1

To remove the discontinuity at
`\displaystyle\sf x=1`, we need to find the new value
`\displaystyle\sf f(1)` should be assigned.

Given the function
`\displaystyle\sf f(x)`:

`\displaystyle\sf f(x)=\begin{cases}x^(2)+2, & x<1\\2x, & x=1\\3, & x>1\end{cases}`

To remove the discontinuity at
`\displaystyle\sf x=1`, we need to ensure that the left-hand limit and the right-hand limit of
`\displaystyle\sf f(x)` at
`\displaystyle\sf x=1` are equal.

The left-hand limit is obtained by evaluating
`\displaystyle\sf f(x)` as
`\displaystyle\sf x` approaches
`\displaystyle\sf 1` from the left:

`\displaystyle\sf \lim_(x\to 1^(-))f(x)=\lim_(x\to 1^(-))(x^(2)+2)=(1^(2)+2)=3`

The right-hand limit is obtained by evaluating
`\displaystyle\sf f(x)` as
`\displaystyle\sf x` approaches
`\displaystyle\sf 1` from the right:

`\displaystyle\sf \lim_(x\to 1^(+))f(x)=\lim_(x\to 1^(+))3=3`

Since the left-hand limit and the right-hand limit are both equal to
`\displaystyle\sf 3`, we can assign
`\displaystyle\sf f(1)` the value of
`\displaystyle\sf 3` to remove the discontinuity.

Therefore, the new value for
`\displaystyle\sf f(1)` should be
`\displaystyle\sf 3`.

`\huge{\mathfrak{\colorbox{black}{\textcolor{lime}{I\:hope\:this\:helps\:!\:\:}}}}`

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