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Sagot :
Answer:
c is the correct option
Step-by-step explanation:
from,
f'(x) = h >0 f(x + h) - f(x)
h
f(x) = - √2x
f(x + h) = - √(2x + h)
f'(x) = h>0 -√(2x + h) - √2x
h
rationalize the denominator
= h>0 -√(2x + h) + √2x (-√(2x + h) - √2x)
h (-√(2x + h) - √2x)
= h>0 4x + 2h - 4x
h(-√(2x + h) -√2x)
= h>0 2h
h(-√(2x+h) - √2x)
= h>0 2
-√(2x+h) - √2x
[tex]\\ \sf\longmapsto f(x)=\sqrt{2x}[/tex]
Now
[tex]\\ \sf\longmapsto \dfrac{1}{-(\sqrt{2x+2h}-\sqrt{2x})}[/tex]
There we plot 2h because if we break root over then it becomes √2h which satisfies f(x)
[tex]\\ \sf\longmapsto \dfrac{1}{-\sqrt{2x+2h}+\sqrt{2x}}[/tex]
Option D
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