Given the following statements about a function f: R → R, select the right option: P: If f(x) is continuous at x = x0, then it is also differentiable at x = x0. Q: If f(x) is continuous at x = x0, then it may not be differentiable at x = x0. R: If f(x) is differentiable at x = x0, then it is also continuous at x = x0.
P is false, Q is true, R is true
Understanding Function Continuity and Differentiability
In calculus, understanding the properties of functions, such as continuity and differentiability, is fundamental. These concepts describe the behavior of a function at a specific point or over an interval. A function is said to be continuous at a point if its graph can be drawn without lifting the pen, meaning there are no breaks, jumps, or holes at that point. A function is differentiable at a point if it has a well-defined derivative at that point, which geometrically means the function has a unique tangent line at that point and its graph is "smooth" with no sharp corners or cusps.
Analyzing Statement P: Continuous Implies Differentiable?
Statement P says: "If f(x) is continuous at x = x0, then it is also differentiable at x = x0."
This statement is false.
Continuity is a necessary condition for differentiability, but it is not sufficient. This means that while a differentiable function must be continuous, a continuous function is not necessarily differentiable. A classic counterexample is the absolute value function.
- Example: Consider the function \(f(x) = |x|\) at \(x_0 = 0\).
- This function is continuous at \(x = 0\) because \(\lim_{x \to 0} |x| = 0 = f(0)\).
- However, \(f(x) = |x|\) is not differentiable at \(x = 0\). The graph of \(|x|\) has a sharp corner (a cusp) at \(x = 0\).
- The left-hand derivative at \(x=0\) is \(\lim_{h \to 0^-} \frac{|0+h| - |0|}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1\).
- The right-hand derivative at \(x=0\) is \(\lim_{h \to 0^+} \frac{|0+h| - |0|}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1\).
- Since the left-hand derivative (\(-1\)) is not equal to the right-hand derivative (\(1\)), the derivative does not exist at \(x = 0\).
Therefore, Statement P is incorrect.
Analyzing Statement Q: Continuous May Not Be Differentiable
Statement Q says: "If f(x) is continuous at x = x0, then it may not be differentiable at x = x0."
This statement is true.
As discussed in the analysis of Statement P, there are indeed functions that are continuous at a point but not differentiable at that same point. The example of \(f(x) = |x|\) at \(x_0 = 0\) clearly demonstrates this possibility.
- The function \(f(x) = |x|\) is continuous at \(x = 0\).
- But, it is not differentiable at \(x = 0\) due to the sharp corner.
This statement acknowledges that continuity does not guarantee differentiability, which is a correct understanding of the relationship between these two properties.
Therefore, Statement Q is correct.
Analyzing Statement R: Differentiable Implies Continuous
Statement R says: "If f(x) is differentiable at x = x0, then it is also continuous at x = x0."
This statement is true.
Differentiability at a point implies continuity at that point. If a function is differentiable at \(x_0\), it means that the limit defining the derivative exists:
\[f'(x_0) = \lim_{h \to 0} \frac{f(x_0 + h) - f(x_0)}{h}\]
For this limit to exist and be finite, the numerator \(f(x_0 + h) - f(x_0)\) must approach \(0\) as \(h \to 0\). This implies:
\[\lim_{h \to 0} (f(x_0 + h) - f(x_0)) = 0\]
Which simplifies to:
\[\lim_{h \to 0} f(x_0 + h) = f(x_0)\]
This is precisely the definition of continuity at \(x_0\).
Intuitively, if a function is "smooth" enough to have a well-defined tangent at every point (differentiable), it must not have any breaks or jumps (continuous).
Therefore, Statement R is correct.
Conclusion on Function Properties
Based on the detailed analysis of each statement:
- Statement P: If f(x) is continuous at x = x0, then it is also differentiable at x = x0. (False)
- Statement Q: If f(x) is continuous at x = x0, then it may not be differentiable at x = x0. (True)
- Statement R: If f(x) is differentiable at x = x0, then it is also continuous at x = x0. (True)
Thus, the combination of truth values is P is false, Q is true, R is true.
| Statement | Truth Value | Reason |
|---|---|---|
| P: Continuous \(\Rightarrow\) Differentiable | False | Counterexample: \(f(x) = |x|\) at \(x=0\) is continuous but not differentiable. |
| Q: Continuous \(\Rightarrow\) Not necessarily Differentiable | True | Acknowledges the possibility shown by counterexamples like \(f(x) = |x|\) at \(x=0\). |
| R: Differentiable \(\Rightarrow\) Continuous | True | The existence of the derivative implies the limit definition of continuity is met. |
Important Questions from Differentiability
What is the value of f'(x) at x = 4 from the following table of values?
x 1 2 3 4 f(x) 20 22 27 35 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:
The set of all point where the function f(x) = 2x|x| is differentiable, is:
