Find the value of k that makes f(x) continuous at x=-1 f(x)= ) -2x^2-2x&ifxleqslant -1 -kx+7&ifxgt -1 k= square

To find the value of \( k \) that makes \( f(x) \) continuous at \( x = -1 \), we need to ensure that the left-hand limit and the right-hand limit of \( f(x) \) at \( x = -1 \) are equal, and both equal to \( f(-1) \).<br /><br />First, let's find \( f(-1) \):<br />\[ f(x) = -2x^2 - 2x \quad \text} \quad x \leq -1 \]<br />\[ f(-1) = -2(-1)^2 - 2(-1) = -2(1) + 2 = -2 + 2 = 0 \]<br /><br />Next, we find the left-hand limit as \( x \) approaches -1:<br />\[ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (-2x^2 - 2x) = -2(-1)^2 - 2(-1) = -2(1) + 2 = -2 + 2 = 0 \]<br /><br />Now, we find the right-hand limit as \( x \) approaches -1:<br />\[ \lim_{x \to -1^+} f(x) = \lim_{x \to^+} (-kx + 7) = -k(-1) + 7 = k + 7 \]<br /><br />For \( f(x) \) to be continuous at \( x = -1 \), these limits must be equal:<br />\[ \lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x) \]<br />\[ 0 = k + 7 \]<br /><br />Solving for \( k \):<br />\[ k + 7 = 0 \]<br />\[ k = -7 \]<br /><br />Therefore, the value of \( k \) that makes \( f(x) \) continuous at \( x = -1 \) is:<br />\[ k = -7 \]