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Slope of the tangent line via the limit definition

Let f(x)=x2+3x−1f(x) = x^2 + 3x - 1.

Use the limit definition of the derivative, f′(x)=lim⁡h→0f(x+h)−f(x)h,f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}, to find f′(1)f'(1), the slope of the tangent line to f(x)=x2+3x−1f(x) = x^2 + 3x - 1 at x=1x = 1. Show the setup of the difference quotient before taking the limit.

Hint

First compute f(1+h)f(1+h) and f(1)f(1) explicitly, then form the difference quotient f(1+h)−f(1)h\frac{f(1+h)-f(1)}{h}, simplify by expanding and canceling hh, and finally let h→0h \to 0.

Step-by-step solution

Compute f(1+h)=(1+h)2+3(1+h)−1=1+2h+h2+3+3h−1=h2+5h+3f(1+h) = (1+h)^2 + 3(1+h) - 1 = 1 + 2h + h^2 + 3 + 3h - 1 = h^2 + 5h + 3. Also, f(1)=12+3(1)−1=3f(1) = 1^2 + 3(1) - 1 = 3. So the difference quotient is f(1+h)−f(1)h=h2+5h+3−3h=h2+5hh.\frac{f(1+h)-f(1)}{h} = \frac{h^2 + 5h + 3 - 3}{h} = \frac{h^2+5h}{h}.

Factor the numerator to get h(h+5)h=h+5\frac{h(h+5)}{h} = h + 5 for h≠0h \neq 0. Taking the limit as h→0h \to 0 gives f′(1)=lim⁡h→0(h+5)=5f'(1) = \lim_{h\to 0} (h+5) = 5. Thus the slope of the tangent line to f(x)f(x) at x=1x=1 is 55.

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