For what value of the constant c is the function f continuous on (−∞,+∞)? f(x)={(cx2)+4x,(x3)−cx}ifx&lt;5,ifx⇒5 c=

Given: f(x)={(cx2)+4x,(x3)−cx}ifx&lt;5,ifx⇒5 The function is continuous on (−∞,+∞) when limx→5−f(x)=f(5)lim f(x)=f(5). That is, the two parts of the function approach the same point at x=5. Let us evaluate both parts of the definition of f at x=5: f(5)=(53)−c(5)=125−5c lim(x→5)f(x)=c(52)+4(5)=25c+20 Since we require limx→5−f(x)=f(5): 25c+20=125−5c Add 5c5c to each side of the previous equation: 30c+20=125 Subtract 20 from each side of the previous equation: 30c=105 Divide each side of the previous equation by 30: c=10530 Evaluate: c=72=3.5

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pedzenekO

Answered question

2021-01-06

For what value of the constant c is the function f continuous on $(- \infty, + \infty) ?$
$f (x) = {(c x^{2}) + 4 x, (x^{3}) - c x} if x < 5, if x \Rightarrow 5$ c=

Answer & Explanation

Clara Reese

Skilled2021-01-07Added 120 answers

Given:
$f (x) = {(c x^{2}) + 4 x, (x^{3}) - c x} if x < 5, if x \Rightarrow 5$
The function is continuous on $(- \infty, + \infty)$ when $lim x \to 5 - f (x) = f (5) lim$ f(x)=f(5). That is, the two parts of the function approach the same point at x=5.
Let us evaluate both parts of the definition of f at x=5: $f (5) = (5^{3}) - c (5) = 125 - 5 c$
$lim (x \to 5) f (x) = c (5^{2}) + 4 (5) = 25 c + 20$
Since we require $lim x \to 5 - f (x) = f (5) :$
25c+20=125−5c
Add 5c5c to each side of the previous equation: 30c+20=125
Subtract 20 from each side of the previous equation: 30c=105
Divide each side of the previous equation by 30: $c = \frac{105}{30}$
Evaluate: $c = \frac{7}{2} = 3.5$

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