# Use the four-step process to find f'(x), then f'(1), f'2), f'(3) f(x)=8+7√x

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### 1 Answer

given f(x) = 8 + 7`sqrt(x)`

and f'(x) = lim (h-->0) (f(x+h)-f(x))/h

**Step 1**: finding f(x+h).

f(x+h) = 8+7`sqrt(x+h)`

**Step 2**: find f(x+h) - f(x)

f(x+h)-f(x) = 8+7`sqrt(x+h)` - (8+7`sqrt(x)` )= 8+7` ` - 8-7`sqrt(x)` = 7 (`sqrt(x+h)-sqrt(x)` )

It can be further simplified by multiplying and dividing by (`sqrt(x+h)+sqrt(x)` ) and using the relation (a+b)(a-b) = a^2-b^2

simplifying, f(x+h) - f(x) = 7(x+h-x)/(`sqrt(x+h)+sqrt(x)` )=7h/(`sqrt(x+h)+sqrt(x)` )

**Step 3**: find (1/h) x f(x+h)-f(x) = (1/h) x 7h/(`sqrt(x+h)+sqrt(x)` ) = 7/(`sqrt(x+h)+sqrt(x)` )

**Step 4**: find lim(h-->0) (1/h)x f(x+h)-f(x) = lim(h-->0) 7/(`sqrt(x+h)+sqrt(x)` )= 7/(2`sqrt(x))`

Thus f'(x) = 7/(2`sqrt(x)` )

Therefore, f'(1) = 7/2

f'(2) = 7/(2`sqrt(2)` )

and f'(3) = 7/(2`sqrt(3)` )

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