For what value of the constant c is the function f continuous on

Armorikam

Armorikam

Answered question

2021-09-12

For what value of the constant c is the function f continuous on (, )?
f(x)=cx2+2xx21 if x<2 and x3cx if x2

Answer & Explanation

crocolylec

crocolylec

Skilled2021-09-13Added 100 answers

f(x)={cx2+2xx<2x3cxx2
Note that: cx2+2x and x3cx are polynomials and are always continuous.
We only need to check for the continuity at x=2
f(x) will be continuous at x=2, if and only if
limx2f(x)=limx2+f(x)
limx2cx2+2x=limx2+x3cx
c22+22=23c26c=4
c=23
Result:
c=23

xleb123

xleb123

Skilled2023-05-09Added 181 answers

Result: 23
Solution:
For the function f(x) to be continuous on (,), we need to ensure that the left-hand and right-hand pieces of the function match at x=2. This means that the limit of f(x) as x approaches 2 from both sides should be equal.
First, let's consider the left-hand piece of the function for x<2:
f(x)=cx2+2xx21
To find the limit as x approaches 2 from the left, we substitute x=2 into the expression:
limx2f(x)=c(22)+2(2)221
Simplifying further:
limx2f(x)=4c+43
Next, let's consider the right-hand piece of the function for x2:
f(x)=x3cx
To find the value of f(x) at x=2, we substitute x=2 into the expression:
f(2)=23c(2)=82c
To ensure continuity at x=2, the limit and the function value must be equal. Therefore, we have:
limx2f(x)=f(2)
Substituting the expressions we found earlier:
4c+43=82c
To solve this equation for c, we can multiply both sides by 3 to eliminate the fraction:
4c+4=246c
Combining like terms:
10c=20
Dividing both sides by 10:
c=2010=21=23
Hence, the constant c that makes the function f(x) continuous on (,) is c=23.
alenahelenash

alenahelenash

Expert2023-05-09Added 556 answers

To find the value of the constant (c) for which the function (f) is continuous on (,), we need to ensure that the left-hand limit and right-hand limit of (f(x)) as (x) approaches 2 from both sides are equal.
First, let's calculate the left-hand limit of (f(x)) as (x) approaches 2:
[limx2f(x)=limx2cx2+2xx21]
Substituting (x = 2) into the expression, we have:
[limx2f(x)=c(2)2+2(2)(2)21=4c+441=4c+43]
Now, let's calculate the right-hand limit of (f(x)) as (x) approaches 2:
[limx2+f(x)=limx2+(x3cx)]
Substituting (x = 2) into the expression, we have:
[limx2+f(x)=(2)3c(2)=82c]
For the function (f(x)) to be continuous at (x = 2), the left-hand limit and right-hand limit must be equal. Therefore, we have the equation:
[4c+43=82c]
To solve for (c), we can start by multiplying both sides of the equation by 3 to eliminate the fraction:
[4c+4=3(82c)]
Next, distribute 3 on the right-hand side:
[4c+4=246c]
Now, let's simplify the equation by combining like terms:
[4c+6c=244]
[10c=20]
Finally, divide both sides of the equation by 10 to solve for (c):
[c=2010=2]
Hence, the value of the constant (c) for which the function (f) is continuous on (,) is (c=23).

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