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If f(x) is differentiable everywhere, then
- a)| f | is differentiable everywhere
- b)| f |2 is differentiable everywhere
- c)f | f | is not differentiable at some point
- d)f + | f | is differentiable everywhere
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
If f(x) is differentiable everywhere, thena)| f | is differentiable ev...
Even if f(x) is differentiable everywhere, |f(x)| need not be differentiable everywhere. An example being where the function changes its sign from negative to positive. f(x)=x at x=0B.
∣f∣= {f if f>0 −f if f<0}
So,∣f∣2 = {f2 if f>0 −f2 if f<0}
Differentiating ∣f∣2,
{2ff′ if f>0 −2ff′ if f<0}
At f=0, LHD=RHD=∣f(0)∣2=0, so function is differentiable.
∣f∣= {f if f>0 −f if f<0}
So,∣f∣2 = {f2 if f>0 −f2 if f<0}
Differentiating ∣f∣2,
{2ff′ if f>0 −2ff′ if f<0}
At f=0, LHD=RHD=∣f(0)∣2=0, so function is differentiable.
Most Upvoted Answer
If f(x) is differentiable everywhere, thena)| f | is differentiable ev...
Differentiability of |f|²
To understand why option B is correct, let's consider the function |f|².
Differentiability of |f|
First, let's discuss the differentiability of |f|. The absolute value function, denoted as |x|, is defined as:
|x| = x, if x ≥ 0
|x| = -x, if x < />
Now, if f is a differentiable function, it means that the derivative of f exists at every point in its domain.
The absolute value function is not differentiable at x = 0 because it has a sharp corner at that point. The derivative of |x| is not defined at x = 0.
Therefore, if we consider the function |f|, it will not be differentiable at points where f = 0 because the absolute value function is not differentiable at those points.
Differentiability of |f|²
Now, let's consider the function |f|². This function is obtained by squaring the absolute value of f.
|f|² = (|f|)²
The square of any real number is always non-negative. Therefore, |f|² will always be non-negative.
Since |f|² is non-negative everywhere, it does not have any sharp corners or points where it is not defined.
The derivative of |f|² can be calculated using the chain rule of differentiation. If g(x) = f(x)², then the derivative of g(x) with respect to x is:
g'(x) = 2f(x)f'(x)
Since f(x) is assumed to be differentiable everywhere, f'(x) exists at every point in the domain of f. Therefore, g(x) = |f|² is also differentiable everywhere.
Conclusion
Based on the above discussion, we can conclude that option B is correct. The function |f|² is differentiable everywhere, whereas the other options are either not differentiable at certain points or not differentiable everywhere.
To understand why option B is correct, let's consider the function |f|².
Differentiability of |f|
First, let's discuss the differentiability of |f|. The absolute value function, denoted as |x|, is defined as:
|x| = x, if x ≥ 0
|x| = -x, if x < />
Now, if f is a differentiable function, it means that the derivative of f exists at every point in its domain.
The absolute value function is not differentiable at x = 0 because it has a sharp corner at that point. The derivative of |x| is not defined at x = 0.
Therefore, if we consider the function |f|, it will not be differentiable at points where f = 0 because the absolute value function is not differentiable at those points.
Differentiability of |f|²
Now, let's consider the function |f|². This function is obtained by squaring the absolute value of f.
|f|² = (|f|)²
The square of any real number is always non-negative. Therefore, |f|² will always be non-negative.
Since |f|² is non-negative everywhere, it does not have any sharp corners or points where it is not defined.
The derivative of |f|² can be calculated using the chain rule of differentiation. If g(x) = f(x)², then the derivative of g(x) with respect to x is:
g'(x) = 2f(x)f'(x)
Since f(x) is assumed to be differentiable everywhere, f'(x) exists at every point in the domain of f. Therefore, g(x) = |f|² is also differentiable everywhere.
Conclusion
Based on the above discussion, we can conclude that option B is correct. The function |f|² is differentiable everywhere, whereas the other options are either not differentiable at certain points or not differentiable everywhere.
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If f(x) is differentiable everywhere, thena)| f | is differentiable everywhereb)| f |2is differentiable everywherec)f | f | is not differentiable at some pointd)f + | f | is differentiable everywhereCorrect answer is option 'B'. Can you explain this answer?
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