Limits

Step 1: Factorizing the numerator and denominator

Step 2: Cancel common terms For , canceling ,

Step 3: Substitute

Step 1: Use the point (2, 3) on the curve y² = ax³ + b.
Since (2, 3) lies on the curve y² = ax³ + b, we have: 3² = a(2³) + b
9 = 8a + b ...(1)

Step 2: Differentiate the equation y² = ax³ + b with respect to x.
Differentiating both sides with respect to x, we get: 2y (dy/dx) = 3ax²
dy/dx = (3ax²) / (2y)

Step 3: Find the slope of the tangent at (2, 3).
The slope of the tangent at (2, 3) is given by: (dy/dx) |_{(2, 3)} = (3a(2²)) / (2(3)) = (12a) / 6 = 2a

Step 4: Compare the slope with the given tangent equation.
The given tangent equation is y = 4x - 5.
The slope of this tangent is 4.
Therefore, 2a = 4, so a = 2.

Step 5: Substitute the value of a in equation (1) to find b.
Substitute a = 2 in 9 = 8a + b: 9 = 8(2) + b
9 = 16 + b
b = 9 - 16 = -7

Step 6: Calculate a² + b².
a² + b² = 2² + (-7)² = 4 + 49 = 53

Therefore, the value of a² + b² is 53.

Step 1: Verify the conditions for Rolle's theorem.
The function f(x) = x(x+3)e^-x/2 is continuous on [-3, 0] and differentiable on (-3, 0) since it is a product of polynomial and exponential functions.
Also, f(-3) = (-3)(-3+3)e^-(-3)/2 = (-3)(0)e^3/2 = 0
and f(0) = 0(0+3)e^-0/2 = 0(3)e⁰ = 0.
Since f(-3) = f(0) = 0, Rolle's theorem is applicable.

Step 2: Find the derivative of f(x).
f(x) = (x² + 3x)e^-x/2
f'(x) = (2x + 3)e^-x/2 + (x² + 3x)e^-x/2(-1/2)
f'(x) = e^-x/2(2x + 3 - (x² + 3x)/2)
f'(x) = e^-x/2((4x + 6 - x² - 3x)/2)
f'(x) = (e^-x/2/2)(-x² + x + 6)

Step 3: Set f'(c) = 0 and solve for c.
By Rolle's theorem, there exists a c (-3, 0) such that f'(c) = 0.
(e^-c/2/2)(-c² + c + 6) = 0
Since e^-c/2 0, we have -c² + c + 6 = 0.
c² - c - 6 = 0
(c - 3)(c + 2) = 0
c = 3 or c = -2.
Since c (-3, 0), we have c = -2.
Therefore, the value of c is -2.

We are given the function defined piecewise. Let's examine the continuity of the function.

Step 1: Check if the function is continuous at . For to be continuous at , the left-hand and right-hand limits must be equal to the value at . The left-hand limit: The right-hand limit: Since the left-hand and right-hand limits are not equal, the function is not continuous at .

Step 2: Analyze the maximum and minimum values. Since takes the value 2 at and the absolute values of the function outside of 0 never exceed 2, the maximum value is 2. Thus, the function has a maximum value of 2.