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The function ๐‘“(๐‘ฅ)=๐‘ฅ2โˆ’9๐‘ฅโˆ’3 can be made continuous at ๐‘ฅ=3 by defining ๐‘“(3) to be:

Question

The function ๐‘“(๐‘ฅ)=๐‘ฅ2โˆ’9๐‘ฅโˆ’3 can be made continuous at ๐‘ฅ=3 by defining ๐‘“(3) to be:

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Solution

The function f(x) = x^2 - 9x - 3 is a polynomial function, and polynomial functions are continuous everywhere in their domain. Therefore, the function is already continuous at x = 3. The value of the function at x = 3 is f(3) = 3^2 - 9*3 - 3 = 9 - 27 - 3 = -21. So, f(3) is already defined and equals -21.

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The function ๐‘“(๐‘ฅ)=๐‘ฅ2โˆ’9๐‘ฅโˆ’3 can be made continuous at ๐‘ฅ=3 by defining ๐‘“(3) to be:Group of answer choices96-6-9

Find the value of the constant ๐‘Ž that will make this piecewise function continuous everywhere.๐‘“(๐‘ฅ)=โŽงโŽฉโŽจโŽชโŽช๐‘ฅ+1๐‘Ž1โˆ’๐‘ฅ๐‘ฅ<0๐‘ฅ=0๐‘ฅ>0

Which function has a discontinuity at x=3?Responsesf(x)={3x+1ย forย x<3x2+1ย forย xโ‰ฅ3๐‘“(๐‘ฅ)={3๐‘ฅ+1ย ๐‘“๐‘œ๐‘Ÿย ๐‘ฅ<3๐‘ฅ2+1ย ๐‘“๐‘œ๐‘Ÿย ๐‘ฅโ‰ฅ3f(x)={3x+1ย forย x<3x2+1ย forย xโ‰ฅ3๐‘“(๐‘ฅ)={3๐‘ฅ+1ย ๐‘“๐‘œ๐‘Ÿย ๐‘ฅ<3๐‘ฅ2+1ย ๐‘“๐‘œ๐‘Ÿย ๐‘ฅโ‰ฅ3f(x)=|xโˆ’3|+2๐‘“(๐‘ฅ)=|๐‘ฅโˆ’3|+2f of x is equal to start absolute value x minus 3 end absolute value plus 2f(x)=xโˆ’3x2๐‘“(๐‘ฅ)=๐‘ฅโˆ’3๐‘ฅ2f of x is equal to the fraction with numerator x minus 3 and denominator x squaredf(x)=x+2x2โˆ’9

Determine the continuity of the function ๐‘“๐‘ฅ=๐‘ฅ3-๐‘ฅ

ist the value(s) of ๐‘ฅ at which ๐‘“ is discontinuous.

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