To find the value of $c$ that makes the function $f(x)$ continuous, we need to ensure that the function does not have any jump discontinuities at $x = 1$. Specifically, the left-hand limit and right-hand limit of $f(x)$ at $x = 1$ should be equal to the function's value at $x = 1$.

Given:

2x^2 + 3cx + 5 & \text{for } x < 1, \\ cx^2 + 3x - 2 & \text{for } x \geq 1. \end{cases}$$ ### Step 1: Calculate the left-hand limit as $$x$$ approaches 1 The left-hand limit of $$f(x)$$ as $$x$$ approaches 1 is: $$\lim_{x \to 1^-} f(x) = 2(1)^2 + 3c(1) + 5 = 2 + 3c + 5 = 3c + 7.$$ ### Step 2: Calculate the right-hand limit as $$x$$ approaches 1 The right-hand limit of $$f(x)$$ as $$x$$ approaches 1 is: $$\lim_{x \to 1^+} f(x) = c(1)^2 + 3(1) - 2 = c + 3 - 2 = c + 1.$$ ### Step 3: Set the left-hand limit equal to the right-hand limit For $$f(x)$$ to be continuous at $$x = 1$$, we set the left-hand limit equal to the right-hand limit: $$3c + 7 = c + 1.$$ ### Step 4: Solve for $$c$$ To solve for $$c$$, subtract $$c$$ from both sides: $$2c + 7 = 1.$$ Now, subtract 7 from both sides: $$2c = -6.$$ Finally, divide by 2: $$c = -3.$$ ### Conclusion: The value of $$c$$ that makes the function $$f(x)$$ continuous at $$x = 1$$ is $$c = -3$$. Would you like further details or have any questions? --- Here are 5 related questions: 1. How do you determine if a function is continuous at a specific point? 2. What is the significance of matching left-hand and right-hand limits in continuity problems? 3. How would the solution change if the piecewise function had a different form? 4. Can you have a situation where $$c$$ could result in discontinuity? What would that look like? 5. How can you check continuity for more complex piecewise functions involving trigonometric functions? **Tip:** Always remember to check the function's value at the point where you're testing for continuity, as it must match both the left-hand and right-hand limits.