Math

QuestionCalculate the limit f(x+h)f(x)h\frac{f(x+h)-f(x)}{h} for h0h \neq 0 where f(x)=x2+5x1f(x)=x^{2}+5x-1 and f(x)=1x+3f(x)=\frac{1}{x+3}.

Studdy Solution

STEP 1

Assumptions1. The function f(x)f(x) is given as f(x)=x+5x1f(x)=x^{}+5 x-1 for the first part of the problem. . The function f(x)f(x) is given as f(x)=1x+3f(x)=\frac{1}{x+3} for the second part of the problem.
3. We are asked to find the limit f(x+h)f(x)h\frac{f(x+h)-f(x)}{h}, where h0h \neq0.

STEP 2

First, let's solve for the first function f(x)=x2+5x1f(x)=x^{2}+5 x-1. We need to substitute x+hx+h into the function f(x)f(x).
f(x+h)=(x+h)2+5(x+h)1f(x+h) = (x+h)^{2}+5 (x+h)-1

STEP 3

Expand the expression f(x+h)f(x+h).
f(x+h)=x2+2xh+h2+5x+5h1f(x+h) = x^{2}+2xh+h^{2}+5x+5h-1

STEP 4

Now, calculate f(x+h)f(x)f(x+h)-f(x).
f(x+h)f(x)=x2+2xh+h2+x+h1(x2+x1)f(x+h)-f(x) = x^{2}+2xh+h^{2}+x+h-1 - (x^{2}+x-1)

STEP 5

implify the expression f(x+h)f(x)f(x+h)-f(x).
f(x+h)f(x)=2xh+h2+5hf(x+h)-f(x) =2xh+h^{2}+5h

STEP 6

Now, divide the expression f(x+h)f(x)f(x+h)-f(x) by hh.
f(x+h)f(x)h=2xh+h2+5hh\frac{f(x+h)-f(x)}{h} = \frac{2xh+h^{2}+5h}{h}

STEP 7

implify the expression f(x+h)f(x)h\frac{f(x+h)-f(x)}{h}.
f(x+h)f(x)h=2x+h+5\frac{f(x+h)-f(x)}{h} =2x+h+5Now, let's move on to the second function f(x)=1x+3f(x)=\frac{1}{x+3}.

STEP 8

Substitute x+hx+h into the function f(x)f(x).
f(x+h)=1x+h+3f(x+h) = \frac{1}{x+h+3}

STEP 9

Now, calculate f(x+h)f(x)f(x+h)-f(x).
f(x+h)f(x)=x+h+3x+3f(x+h)-f(x) = \frac{}{x+h+3} - \frac{}{x+3}

STEP 10

To simplify this expression, find a common denominator.
f(x+h)f(x)=(x+3)(x+h+3)(x+h+3)(x+3)f(x+h)-f(x) = \frac{(x+3)-(x+h+3)}{(x+h+3)(x+3)}

STEP 11

implify the expression f(x+h)f(x)f(x+h)-f(x).
f(x+h)f(x)=h(x+h+3)(x+3)f(x+h)-f(x) = \frac{-h}{(x+h+3)(x+3)}

STEP 12

Now, divide the expression f(x+h)f(x)f(x+h)-f(x) by hh.
f(x+h)f(x)h=hh(x+h+)(x+)\frac{f(x+h)-f(x)}{h} = \frac{-h}{h(x+h+)(x+)}

STEP 13

implify the expression f(x+h)f(x)h\frac{f(x+h)-f(x)}{h}.
f(x+h)f(x)h=(x+h+3)(x+3)\frac{f(x+h)-f(x)}{h} = \frac{-}{(x+h+3)(x+3)}So, for the function f(x)=x2+5xf(x)=x^{2}+5 x-, f(x+h)f(x)h=2x+h+5\frac{f(x+h)-f(x)}{h} =2x+h+5, and for the function f(x)=x+3f(x)=\frac{}{x+3}, f(x+h)f(x)h=(x+h+3)(x+3)\frac{f(x+h)-f(x)}{h} = \frac{-}{(x+h+3)(x+3)}.

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