If the function \(f\left( x \right) = \left\{ {\begin{array}{c} {a{x^2} - b,}&{\left| x \right| < 1}\\ { - \frac{1}{{\left| x \right|}},}&{\left| x \right| \ge 1} \end{array}} \right.\) is differentiable, then the values of a and b are respectively:
0.5, 1.5
Finding Parameters for a Differentiable Piecewise Function
The given function is defined as:
\(f\left( x \right) = \left\{ {\begin{array}{c} {a{x^2} - b,}&{\left| x \right| < 1}\\ { - \frac{1}{{\left| x \right|}},}&{\left| x \right| \ge 1} \end{array}} \right.\)
This can be rewritten by considering the absolute value:
\(f\left( x \right) = \left\{ {\begin{array}{c} {a{x^2} - b,}&{ - 1 < x < 1}\\ { - \frac{1}{{ - x}} = \frac{1}{x},}&{ x \le - 1}\\ { - \frac{1}{x},}&{ x \ge 1} \end{array}} \right.\)
For the function \(f(x)\) to be differentiable everywhere, it must first be continuous everywhere. The potential points where continuity and differentiability might fail are where the definition of the function changes, which are \(x = 1\) and \(x = -1\).
Ensuring Continuity of the Piecewise Function
For continuity at a point, the left-hand limit, the right-hand limit, and the function value at that point must be equal.
At \(x = 1\):
- Left-hand limit: \(\mathop {\lim }\limits_{x \to {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {a{x^2} - b} \right) = a{(1)^2} - b = a - b\)
- Right-hand limit: \(\mathop {\lim }\limits_{x \to {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( { - \frac{1}{x}} \right) = - \frac{1}{1} = - 1\)
- Function value: \(f(1) = - \frac{1}{{\left| 1 \right|}} = - \frac{1}{1} = - 1\)
For continuity at \(x=1\), we must have \(a - b = -1\).
At \(x = -1\):
- Left-hand limit: \(\mathop {\lim }\limits_{x \to - {1^ - }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ - }} \left( {\frac{1}{x}} \right) = \frac{1}{{ - 1}} = - 1\)
- Right-hand limit: \(\mathop {\lim }\limits_{x \to - {1^ + }} f\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ + }} \left( {a{x^2} - b} \right) = a{( - 1)^2} - b = a - b\)
- Function value: \(f(-1) = - \frac{1}{{\left| { - 1} \right|}} = - \frac{1}{1} = - 1\)
For continuity at \(x=-1\), we must have \(a - b = -1\). Both points give the same condition for continuity.
So, our first equation is: \(a - b = -1\) (Equation 1)
Ensuring Differentiability of the Piecewise Function
For differentiability at a point, the left-hand derivative and the right-hand derivative must be equal at that point.
Let's find the derivative of each piece:
- For \(-1 < x < 1\), \(f(x) = ax^2 - b\), so \(f'(x) = 2ax\).
- For \(x > 1\), \(f(x) = - \frac{1}{x}\), so \(f'(x) = - ( - {x^{ - 2}}) = {x^{ - 2}} = \frac{1}{x^2}\).
- For \(x < -1\), \(f(x) = \frac{1}{x}\), so \(f'(x) = - {x^{ - 2}} = - \frac{1}{x^2}\).
Now, let's check differentiability at \(x=1\) and \(x=-1\).
At \(x = 1\):
- Left-hand derivative: \(\mathop {\lim }\limits_{x \to {1^ - }} f'\left( x \right) = \mathop {\lim }\limits_{x \to {1^ - }} \left( {2ax} \right) = 2a(1) = 2a\)
- Right-hand derivative: \(\mathop {\lim }\limits_{x \to {1^ + }} f'\left( x \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {\frac{1}{x^2}} \right) = \frac{1}{{(1)}^2} = 1\)
For differentiability at \(x=1\), we must have \(2a = 1\), which gives \(a = \frac{1}{2} = 0.5\).
At \(x = -1\):
- Left-hand derivative: \(\mathop {\lim }\limits_{x \to - {1^ - }} f'\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ - }} \left( { - \frac{1}{x^2}} \right) = - \frac{1}{{( - 1)}^2} = - 1\)
- Right-hand derivative: \(\mathop {\lim }\limits_{x \to - {1^ + }} f'\left( x \right) = \mathop {\lim }\limits_{x \to - {1^ + }} \left( {2ax} \right) = 2a( - 1) = - 2a\)
For differentiability at \(x=-1\), we must have \(-2a = -1\), which also gives \(a = \frac{{ - 1}}{{ - 2}} = \frac{1}{2} = 0.5\).
Both differentiability conditions give the same value for \(a\): \(a = 0.5\).
Solving for a and b
We have the following system of equations:
- \(a - b = -1\)
- \(a = 0.5\)
Substitute the value of \(a\) from Equation 2 into Equation 1:
\(0.5 - b = -1\)
\( - b = - 1 - 0.5\)
\( - b = - 1.5\)
\(b = 1.5\)
Thus, the values of a and b that make the function differentiable are \(a = 0.5\) and \(b = 1.5\).
Let's summarise the conditions:
| Point | Condition | Requirement | Equation |
|---|---|---|---|
| \(x=1\) | Continuity | \(\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} f(x) = f(1)\) | \(a - b = -1\) |
| \(x=-1\) | Continuity | \(\mathop {\lim }\limits_{x \to -{1^ - }} f(x) = \mathop {\lim }\limits_{x \to -{1^ + }} f(x) = f(-1)\) | \(a - b = -1\) |
| \(x=1\) | Differentiability | \(\mathop {\lim }\limits_{x \to {1^ - }} f'(x) = \mathop {\lim }\limits_{x \to {1^ + }} f'(x)\) | \(2a = 1\) |
| \(x=-1\) | Differentiability | \(\mathop {\lim }\limits_{x \to -{1^ - }} f'(x) = \mathop {\lim }\limits_{x \to -{1^ + }} f'(x)\) | \(-2a = -1\) |
Solving \(2a=1\) or \(-2a=-1\) gives \(a = 0.5\). Substituting \(a=0.5\) into \(a-b=-1\) gives \(0.5 - b = -1\), so \(b = 1.5\).
Revision Table: Key Concepts
| Concept | Explanation | Relevance Here |
|---|---|---|
| Continuity | A function is continuous at a point if the limit exists at that point and is equal to the function value. For piecewise functions, check at boundary points. | Required at \(x=1\) and \(x=-1\), leading to the condition \(a-b=-1\). |
| Differentiability | A function is differentiable at a point if the derivative exists at that point. For piecewise functions, the left and right derivatives must be equal at boundary points. Differentiability implies continuity. | Required at \(x=1\) and \(x=-1\), leading to the condition \(a=0.5\). |
| Piecewise Function | A function defined by multiple sub-functions, each applying to a certain interval of the domain. | The given function is piecewise, requiring analysis at the points where the definition changes. |
| Absolute Value Function | \(|x|\) is defined as \(x\) for \(x \ge 0\) and \(-x\) for \(x < 0\). | Appears in the definition of \(f(x)\) for \(|x| \ge 1\), requiring separate considerations for \(x \ge 1\) and \(x \le -1\). |
Additional Information on Differentiability
For a function \(f(x)\) to be differentiable at a point \(c\), two main conditions must be met:
- Continuity at \(c\): The function must be continuous at \(x=c\). If a function is not continuous at a point, it cannot be differentiable there.
- Existence and Equality of Left and Right Derivatives: The limit of the difference quotient as \(x\) approaches \(c\) from the left must exist (left derivative, \(f'(c^-)\)) and be equal to the limit of the difference quotient as \(x\) approaches \(c\) from the right (right derivative, \(f'(c^+)\)). For piecewise functions, we often calculate the derivatives of the pieces and check if they match at the boundary points.
In this problem, we first established continuity, which gave us a relationship between \(a\) and \(b\). Then, we used the differentiability condition, which allowed us to find the value of \(a\). Finally, we used the value of \(a\) in the continuity equation to find the value of \(b\).
The values \(a=0.5\) and \(b=1.5\) ensure that the piecewise function transitions smoothly (both in value and slope) at \(x=1\) and \(x=-1\).
Important Questions from Differentiability
The set of all points, where the function \({\rm{f}}\left( {\rm{x}} \right) = \sqrt {1 - {{\rm{e}}^{ - {{\rm{x}}^2}}}} \) is differentiable, is
Let f be a differentiable function defined for all x ∈ R such that f(x3) = x5 for all x ∈ R, x ≠ 0. Then the value of \(\dfrac{df}{dx} (8)\) is:
If \(f(x)=\displaystyle\sum_{n-0}^{2k}\left(a_n|x|^n+b_n\ \sin^2x\right)\), where \(a_i^{'}\)s and \(b_i^{'}\)s (0 ≤ i ≤ k) are real constants, then f(x) is:
If 7x3 + 3y3 + 4x2 + 6x = 100, then (dy/dx)(2, 4) is
- \(\rm\frac{X}{(X+2)(X−3)}\) =
