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Katen [24]
3 years ago
13

PLEASE HELP Which expression represents the difference quotient of the function f (x) = negative StartRoot 2 x EndRoot?

Mathematics
2 answers:
natali 33 [55]3 years ago
5 0

Answer:

c is the correct option

Step-by-step explanation:

from,

f'(x) = h >0 <u>f</u><u>(</u><u>x</u><u> </u><u>+</u><u> </u><u>h</u><u>)</u><u> </u><u>-</u><u> </u><u>f</u><u>(</u><u>x</u><u>)</u><u> </u>

h

f(x) = - √2x

f(x + h) = - √(2x + h)

f'(x) = h>0 <u>-</u><u>√(2x + h) - √2x</u>

h

rationalize the denominator

= h>0 <u>-</u><u>√</u><u>(</u><u>2</u><u>x</u><u> </u><u>+</u><u> </u><u>h</u><u>)</u><u> </u><u>+</u><u> </u><u>√</u><u>2</u><u>x</u><u> </u><u> </u><u>(</u><u>-</u><u>√</u><u>(</u><u>2</u><u>x</u><u> </u><u>+</u><u> </u><u>h</u><u>)</u><u> </u><u>-</u><u> </u><u>√</u><u>2</u><u>x</u><u>)</u>

h (-√(2x + h) - √2x)

= h>0 <u>4</u><u>x</u><u> </u><u>+</u><u> </u><u>2</u><u>h</u><u> </u><u>-</u><u> </u><u>4</u><u>x</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

h(-√(2x + h) -√2x)

= h>0 <u>2</u><u>h</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

h(-√(2x+h) - √2x)

= h>0 <u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>2</u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u>

-√(2x+h) - √2x

MrRissso [65]3 years ago
3 0

\\ \sf\longmapsto f(x)=\sqrt{2x}

Now

\\ \sf\longmapsto \dfrac{1}{-(\sqrt{2x+2h}-\sqrt{2x})}

There we plot 2h because if we break root over then it becomes √2h which satisfies f(x)

\\ \sf\longmapsto \dfrac{1}{-\sqrt{2x+2h}+\sqrt{2x}}

Option D

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