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Consider the given function f(x).
x³ + x² + 1, for x ≥ 1
If the function is differentiable everywhere, the value of b must be ______. (rounded off to one decimal place)
The correct answer is -2.0.
For the function to be differentiable at the critical point x=1, it must be both continuous and have equal left-hand and right-hand derivatives. Continuity at x=1 gives the equation a + b = 3. Differentiability at x=1 gives a = 5. Substituting a = 5 into the first equation gives 5 + b = 3, which yields b = -2. Rounded to one decimal place, the answer is -2.0.
For a piecewise function to be differentiable everywhere, it must satisfy two conditions at the point where the function's definition changes (in this case, at x = 1):
- The function must be continuous at that point.
- The function must be "smooth" at that point, meaning the derivative from the left must equal the derivative from the right.
Step 1: Apply the Continuity Condition at x = 1
For f(x) to be continuous at x = 1, the limit as x approaches 1 from the left must equal the limit as x approaches 1 from the right.
Substitute the respective function pieces:
Now, evaluate the limits by plugging in x = 1:
a + b = 1 + 1 + 1
a + b = 3 --- (Equation 1)
Step 2: Apply the Differentiability Condition at x = 1
First, we find the derivative, f'(x), for each piece of the function.
- For x < 1: f'(x) = d/dx (ax + b) = a
- For x > 1: f'(x) = d/dx (x³ + x² + 1) = 3x² + 2x
For the function to be differentiable at x = 1, the left-hand derivative (LHD) must equal the right-hand derivative (RHD) at x = 1.
Evaluate each derivative at x = 1:
a = 3 + 2
a = 5 --- (Equation 2)
Step 3: Solve for b
Now we have a simple system of two equations. Substitute the value of a from Equation 2 into Equation 1:
(5) + b = 3
b = 3 - 5
b = -2
Step 4: Format the Final Answer
The question requires the answer to be rounded off to one decimal place. Therefore, -2 becomes -2.0.
- Thomas, G. B., Weir, M. D., & Hass, J. (2014). Thomas' Calculus. Pearson. (Chapter 3: Differentiation).
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