CCC
Okay, so I need to figure out how to solve this problem. Let me read it again carefully. The problem says:
"Let ( f(x) = \begin{cases}
x^2 & \text{if } x \leq 1 \
ax + b & \text{if } x > 1
\end{cases} )
Find the values of ( a ) and ( b ) such that ( f(x) ) is differentiable at ( x = 1 )."
Alright, so we have a piecewise function. For x less than or equal to 1, it's a quadratic function ( x^2 ), and for x greater than 1, it's a linear function ( ax + b ). The goal is to find the constants a and b that make this function differentiable at x = 1.
First, I remember that for a function to be differentiable at a point, it has to be continuous there first. So I need to ensure that the left-hand limit and the right-hand limit at x = 1 are equal, and also that the derivatives from the left and the right are equal. That gives two conditions to solve for a and b.
Let me start with continuity. If the function is continuous at x = 1, then the value of the function from the left (using ( x^2 )) must equal the value from the right (using ( ax + b )).
Calculating the left-hand limit as x approaches 1: ( \lim_{x \to 1^-} f(x) = (1)^2 = 1 ).
Calculating the right-hand limit as x approaches 1: ( \lim_{x \to 1^+} f(x) = a(1) + b = a + b ).
For continuity, these two must be equal, so:
( a + b = 1 ). That's the first equation.
Now, differentiability. The derivative from the left and the derivative from the right at x = 1 must also be equal.
First, let's find the derivative of the left piece. The derivative of ( x^2 ) is ( 2x ). So, evaluating this at x = 1 gives ( 2(1) = 2 ).
Now, the derivative of the right piece, which is ( ax + b ), is just ( a ). So the derivative from the right as x approaches 1 is ( a ).
For differentiability, these derivatives must be equal:
( a = 2 ).
Now that we have a value for a, we can substitute back into the continuity equation to find b.
From the first equation: ( a + b = 1 ).
Substituting a = 2:
( 2 + b = 1 )
Subtracting 2 from both sides:
( b = 1 - 2 = -1 ).
So, a should be 2 and b should be -1.
Wait, let me double-check that. If a is 2 and b is -1, then for x > 1, the function is 2x - 1. Let's verify continuity at x = 1:
Left-hand limit: 1^2 = 1.
Right-hand limit: 2(1) -1 = 2 -1 = 1. Okay, that matches.
Now check differentiability:
Left derivative at x=1 is 2(1) = 2.
Right derivative is a = 2. So they match. That seems correct.
Hmm, but let me make sure there's no mistake here. Sometimes when piecewise functions are involved, especially with quadratics and lines, you have to ensure that the transition point is smooth.
Another way to check is to write out the function with these values and see if it's differentiable. Let's define f(x) as:
For x ≤1: x²
For x >1: 2x -1
At x=1, both pieces give 1, so it's continuous. The derivative from the left is 2x, which is 2 at x=1. The derivative from the right is 2, which is the same. So yes, differentiable.
Therefore, the values are a=2 and b=-1.
But wait, let me think again. Is there a possibility I missed something? For example, maybe I need to check higher-order derivatives? But the problem only asks for differentiability, not twice differentiable. So as long as the first derivative exists and is continuous, which in this case, since the left and right derivatives match, it's differentiable.
Alternatively, sometimes people might confuse differentiability with continuity, but since we already enforced continuity first, which is a prerequisite for differentiability, and then ensured the derivatives match, that should be sufficient.
Yes, I think that's correct. So the answer is a=2 and b=-1.
Final Answer
The values of ( a ) and ( b ) are (\boxed{a = 2}) and (\boxed{b = -1}).
https://www.18122.com/url/usdt/usdt4259.jpeg
To find the values of ( a ) and ( b ) such that ( f(x) ) is differentiable at ( x = 1 ), we need to ensure that the function is continuous at ( x = 1 ) and that the derivatives from the left and right at ( x = 1 ) are equal.
Continuity at ( x = 1 ):
The left-hand limit as ( x ) approaches 1 is ( \lim_{x \to 1^-} f(x) = 1^2 = 1 ).
The right-hand limit as ( x ) approaches 1 is ( \lim_{x \to 1^+} f(x) = a(1) + b = a + b ).
For continuity, these must be equal: ( a + b = 1 ).
Differentiability at ( x = 1 ):
The derivative of the left piece ( x^2 ) is ( 2x ). Evaluating this at ( x = 1 ) gives ( 2(1) = 2 ).
The derivative of the right piece ( ax + b ) is ( a ).
For differentiability, these derivatives must be equal: ( a = 2 ).
Substituting ( a = 2 ) into the continuity equation ( a + b = 1 ):
[ 2 + b = 1 ]
[ b = 1 - 2 ]
[ b = -1 ]
Thus, the values of ( a ) and ( b ) are (\boxed{a = 2}) and (\boxed{b = -1}).
页:
[1]