If f and g are 2 Non-Differentiable functions at x = a.
S₁:
f+g is always Non-Differentiable at x = a.
S₂:
f.g is always Non-Differentiable at x = a.
S₁:
f+g is always Non-Differentiable at x = a.
S₂:
f.g is always Non-Differentiable at x = a.
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Answers & Comments
Answer:
f
′
(x)=e
(f(x)−g(x))
g
′
(x)∀xϵR
⇒e
−f(x)
⋅f
′
(x)−e
−g(x)
g
′
(x)=0
⇒∫(e
−f(x)
f
′
(x)−e
−g(x)
⋅g
′
(x))dx=C
⇒−e
−f(x)
+e
−g(x)
=C
⇒−e
−f(1)
+e
−g(1)
=−e
−f(2)
+e
−g(2)
⇒−
e
1
+e
−g(1)
=−e
−f(2)
+
e
1
⇒e
−f(2)
+e
−g(1)
=
e
2
∴ e
−f(2)
<
e
2
and e
−g(1)
<
e
2
⇒−f(2)<log
e
2−1 and −g(1)<log
e
2−1
⇒f(2)>1−log
e
2 and g(1)>1−log
e
2.
Answer:
if neither f nor g is differentiable at x=a. is f+g differentiable at x=a?
My answer: yes it is differentiable , because according to this question here: if neither f nor g is differentiable at x=a. is f+g differentiable at x=a?
If f is not differentiable then −f is not differentiable (but I do not understand exactly why ? may be because f is not differentiable meaning that the limit of differentiability does not exist for f and hence it trivially does not exist also for -f ..... am I correct? ). now f+(-f) = 0 which is a differentiable function ....
Step-by-step explanation:
Please mark me as brainliest.