5.5. For what values of $$ k $$ and $$ m $$ is the function continuous everywhere? \[ f(x)=\left\{\begin{array}{ll} x^{2}+5 & \text { if } x2 \ m(x+1)+k & \text { if }-1

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5.5. For what values of $$ k $$ and $$ m $$ is the function continuous everywhere? \[ f(x)=\left\{\begin{array}{ll} x^{2}+5 & \text { if } x>2 \ m(x+1)+k & \text { if }-1

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**Step 1. Identify the boundary points.** The function $$ f(x) $$ is defined by three expressions. The pieces change at $$ x = -1 $$ and $$ x = 2 $$. These are the points where we need to check for continuity. --- **Step 2. Ensure continuity at $$ x =2 $$.** - For $$ x > 2 $$, the function is given by: $$ f(x) = x^2 + 5. $$ Even though this piece is defined for $$ x > 2 $$, coming from the right of $$ x=2 $$ the limit is: $$ \lim_{x \to 2^+} f(x) = 2^2 + 5 = 4 + 5 = 9. $$ - For $$-1 < x \leq 2$$, the function is: $$ f(x) = m(x+1)+k. $$ Since $$ x = 2 $$ belongs to this interval, the value at $$ x = 2 $$ is: $$ f(2) = m(2+1) + k = 3m + k. $$ For continuity at $$ x = 2 $$, the value from the second piece must be equal to the limit from the first piece: $$ 3m + k = 9. $$ We will refer to this as **Equation (1)**. --- **Step 3. Ensure continuity at $$ x = -1 $$.** - For $$ x \leq -1 $$, the function is given by: $$ f(x) = 2x^3 + x + 7. $$ The value at $$ x = -1 $$ is: $$ f(-1) = 2(-1)^3 + (-1) + 7 = 2(-1) - 1 + 7 = -2 - 1 + 7 = 4. $$ - For $$-1 < x \leq 2$$, the function is: $$ f(x) = m(x+1) + k. $$ Although $$ x=-1 $$ is not included in this piece, the right-hand limit as $$ x \to -1^+ $$ is: $$ \lim_{x \to -1^+} f(x) = m(-1+1) + k = m(0) + k = k. $$ For continuity at $$ x = -1 $$, we require: $$ k = 4. $$ This gives us the value of $$ k $$. --- **Step 4. Solve for $$ m $$.** Substitute $$ k = 4 $$ into **Equation (1)**: $$ 3m + 4 = 9. $$ Now solve for $$ m $$: $$ 3m = 9 - 4 = 5, $$ $$ m = \frac{5}{3}. $$ --- **Final Answer:** $$ k = 4 $$ and $$ m = \frac{5}{3} $$. $$ k = 4 $$ and $$ m = \frac{5}{3} $$.

5.5. For what values of and is the function continuous everywhere?

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