Find the value of k that makes f(x) continuous at x=-1 f(x)= ) (2x^3+2x^2)/(x+1)&ifxneq -1 k&ifx=-1 k= square

To find the value of \( k \) that makes \( f(x) \) continuous at \( x = -1 \), we need to ensure that the limit of \( f(x) \) as \( x \) approaches \(-1\) is equal to \( f(-1) \).<br /><br />First, let's find the limit of \( f(x) \) as \( x \) approaches \(-1\). We start by simplifying the expression for \( f(x) \) when \( x \neq -1 \):<br /><br />\[ f(x) = \frac{2x^3 + 2x^2}{x + 1} \]<br /><br />We can factor the numerator:<br /><br />\[ f(x) = \frac{2x^2(x + 1)}{x + 1} \]<br /><br />For \( x \neq -1 \), we can cancel out the \( x + 1 \) terms:<br /><br />\[ f(x) = 2x^2 \]<br /><br />Now, we take the limit of this simplified expression as \( x \) approaches \(-1\):<br /><br />\[ \lim_{x \to -1} f(x) = \lim_{x \to -1} 2x^2 = 2(-1)^2 = 2 \]<br /><br />For \( f(x) \) to be continuous at \( x = -1 \), we need:<br /><br />\[ f(-1) = k = \lim_{x \to -1} f(x) \]<br /><br />Thus, we set \( k \) equal to the limit we found:<br /><br />\[ k = 2 \]<br /><br />Therefore, the value of \( k \) that makes \( f(x) \) continuous at \( x = -1 \) is:<br /><br />\[ k = 2 \]