Question

evaluate
lim{f(2 + h) - f(2)}/{h},
h->0
where f(x) = -3 - sqrt{x}

Answer 1

\(\bf lim_{h \to 0} \cfrac{f(2+h)-f(2)}{h} \qquad \qquad \qquad f(x) = -3-\sqrt{x} \quad ?\)

Answer 2

yes

Answer 3

\(\bf lim_{h \to 0} \cfrac{f(2+h)-f(2)}{h} \qquad \qquad \qquad f(x) = -3-\sqrt{x}\\ \quad \\ \quad \\
f(2) = -3-\sqrt{2} \qquad \qquad f(2+h) = -3-\sqrt{2+h}\\\quad \\
\cfrac{f(2+h)-f(2)}{h} \implies \cfrac{-3-\sqrt{2+h}-(-3-\sqrt{2})}{h}\\\quad \\
\cfrac{\cancel{-3}-\sqrt{2+h}\cancel{+3}+\sqrt{2}}{h} \implies \cfrac{-\sqrt{2+h}+\sqrt{2}}{h} \implies \cfrac{\sqrt{2}-\sqrt{2+h}}{h}\\
\textit{now let's us multiply for the conjugate of the numerator}\\
\cfrac{\sqrt{2}-\sqrt{2+h}}{h} \cdot \cfrac{\sqrt{2}+\sqrt{2+h}}{\sqrt{2}+\sqrt{2+h}} \implies \cfrac{\cancel{2}-(\cancel{2}+\cancel{h})}{\cancel{h}\cdot (\sqrt{2}+\sqrt{2+h})}\)

Answer 4

so it would be \[1/\sqrt{4}=== 1/2\]

Answer 5

\(\bf \sqrt{2} + \sqrt{2} \implies 2\sqrt{2} \)

Answer 6

\(\bf \sqrt{2} \times \sqrt{2} \implies \sqrt{4} \)

Answer 7

so when u add it, one of the square symbols cancels?

Answer 8

well, no, they just add, like you'd with a variable
say \(\bf a = \sqrt{2}\\
a + a = 2a\)

Answer 9

i got it thank, before u go i got one more question.

Answer 10

If f(x) = 19, find f'( 11 )

Answer 11

i tried using f(x+h)−f(x)/h equation but idk

Answer 12

hmmm... f'(x) = 0
dy/dx of a constant is 0

Answer 13

thanks for ur help

Answer 14

yw