Question

for what value of the constant c is the function f continuous on $(-\infty, \infty)$?
$f(x) = \

$$\begin{cases} cx^2 + 8x & \\text{if } x < 5 \\\\ x^3 - cx & \\text{if } x \\geq 5 \\end{cases}$$

$
$c = \square \\ \boldsymbol{\times}$
enhanced feedback
please try again. choose c so that the function is continuous at the point where once piece of the function ends and the other begins.
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Explanation:

Step1: Find left - hand limit at x = 5

For \(x\lt5\), \(f(x)=cx^{2}+8x\). The left - hand limit as \(x\to5^{-}\) is \(\lim_{x\to5^{-}}(cx^{2}+8x)\). Substitute \(x = 5\) into \(cx^{2}+8x\), we get \(c\times5^{2}+8\times5=25c + 40\).

Step2: Find right - hand limit at x = 5

For \(x\geq5\), \(f(x)=x^{3}-cx\). The right - hand limit as \(x\to5^{+}\) is \(\lim_{x\to5^{+}}(x^{3}-cx)\). Substitute \(x = 5\) into \(x^{3}-cx\), we get \(5^{3}-c\times5=125 - 5c\).

Step3: Set left - hand limit equal to right - hand limit (for continuity at x = 5)

Since the function is continuous on \((-\infty,\infty)\), it must be continuous at \(x = 5\). So, \(\lim_{x\to5^{-}}f(x)=\lim_{x\to5^{+}}f(x)\), which means \(25c+40 = 125-5c\).

Step4: Solve the equation for c

Add \(5c\) to both sides of the equation: \(25c + 5c+40=125-5c + 5c\), so \(30c+40 = 125\).
Subtract 40 from both sides: \(30c+40 - 40=125 - 40\), so \(30c=85\).
Divide both sides by 30: \(c=\frac{85}{30}=\frac{17}{6}\approx2.833\). Wait, let's check the calculation again.

Wait, step 3 equation: \(25c + 40=125-5c\)
Add \(5c\) to both sides: \(25c+5c + 40=125\), \(30c+40 = 125\)
Subtract 40: \(30c=125 - 40=85\)? No, 125 - 40 is 85? Wait 125-40 = 85? 125 - 40: 125-40 = 85. Then \(c=\frac{85}{30}=\frac{17}{6}\)? Wait, no, let's re - do the equation:

\(25c+40=125 - 5c\)

Bring all c terms to left and constants to right:

\(25c + 5c=125 - 40\)

\(30c=85\)? Wait, 125 - 40 is 85? 125-40 = 85. But 85 and 30 have a common factor of 5. So \(c=\frac{85\div5}{30\div5}=\frac{17}{6}\approx2.833\). Wait, but let's check the arithmetic again.

Wait, \(x = 5\), left - hand side: \(cx^{2}+8x\) when \(x = 5\) is \(25c+40\). Right - hand side: \(x^{3}-cx\) when \(x = 5\) is \(125-5c\).

Set equal: \(25c + 40=125-5c\)

\(25c+5c=125 - 40\)

\(30c = 85\)? Wait, 125-40 is 85? 125-40: 125 - 40 = 85. Then \(c=\frac{85}{30}=\frac{17}{6}\approx2.833\). Wait, maybe I made a mistake in subtraction. 125 - 40: 125-40 = 85. Yes. So \(c=\frac{17}{6}\) or approximately 2.83.

Wait, let's check with \(c=\frac{17}{6}\):

Left - hand limit at \(x = 5\): \(25\times\frac{17}{6}+40=\frac{425}{6}+\frac{240}{6}=\frac{425 + 240}{6}=\frac{665}{6}\approx110.833\)

Right - hand limit at \(x = 5\): \(125-5\times\frac{17}{6}=125-\frac{85}{6}=\frac{750}{6}-\frac{85}{6}=\frac{750 - 85}{6}=\frac{665}{6}\approx110.833\). So it works.

Wait, but maybe I made a mistake in the equation setup. Let's re - derive the continuity condition. For a piece - wise function to be continuous at a point \(x = a\) where the function changes its definition, we need \(\lim_{x\to a^{-}}f(x)=\lim_{x\to a^{+}}f(x)=f(a)\). Here, at \(x = 5\), \(f(5)=5^{3}-c\times5 = 125 - 5c\) (since for \(x\geq5\), \(f(x)=x^{3}-cx\)). And \(\lim_{x\to5^{-}}f(x)=c\times5^{2}+8\times5=25c + 40\). So setting them equal: \(25c+40=125 - 5c\) is correct.

Solving:

\(25c+5c=125 - 40\)

\(30c=85\)

\(c=\frac{85}{30}=\frac{17}{6}\approx2.833\)

Answer:

\(\frac{17}{6}\) (or approximately \(2.83\))