Find All Values of a for f to be Continuous at c

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The function f is defined as f(x) = 2cos(x) for x ≤ c and f(x) = ax^2 + b for x > c. For f to be continuous at x = c, the condition 2cos(c) = ac^2 + b must hold. If c ≠ 0, the value of a can be expressed as a = (2cos(c) - b)/c^2. However, if c = 0, a solution exists only if b = 2, leading to a discontinuity unless this condition is met.

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A function f is defined as follows:



f(x) = 2cos(x) if x≤c,
= ax^2 + b if x > c .

Where a,b, and c are constants. If b and c are given. find all values of a for which f is continuous at the point x = c

Solution:
a = (2cos(c) - b)/c^2 if c ≠ 0 ; if c = 0 there is no solution unless b = 2.

I don't understand how if c = 0 and b = 2 there is a solution at c = 0

For instance... (2cos(c) - 2)/c^2 = (2/c)( Cos(c) - 1 )/c = 2/0*0 as c → 0 which is still a discontinuity.
 
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In order for f(x) to be continuous at x = 0, f(0) must be defined and:

[tex]\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)[/tex]

Looking at the limit part of this, calculate [itex]\lim_{x \to 0^-} f(x)[/itex]. What do you get? Now look at [itex]\lim_{x \to 0^+} f(x)[/itex]. What is the only value of b that we can have to get the right limit?
 
Miike012 said:
A function f is defined as follows:

f(x) = 2cos(x) if x≤c,
= ax^2 + b if x > c .

Where a,b, and c are constants. If b and c are given. find all values of a for which f is continuous at the point x = c

Solution:
a = (2cos(c) - b)/c^2 if c ≠ 0 ; if c = 0 there is no solution unless b = 2.

I don't understand how if c = 0 and b = 2 there is a solution at c = 0

For instance... (2cos(c) - 2)/c^2 = (2/c)( Cos(c) - 1 )/c = 2/0*0 as c → 0 which is still a discontinuity.
Generalizing on what gb7nash said: f is continuous at at x=c if and only if
[itex]\displaystyle \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = f(c)[/itex]​
Of course since cos(x) is continuous on ℝ, [itex]\displaystyle \lim_{x \to c^-}f(x)=f(c)[/itex] is true for all real c.

[itex]\displaystyle \lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x)[/itex] leads directly to the equation
[itex]\displaystyle 2\cos(c)=ac^2+b[/itex]​
Use that last equation to solve the two cases:
(1) c = 0

(2) c ≠ 0​
 

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