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Use the definition of the derivative at a point to find the slope of the tangent line of f(x)=3x^2+2x+1 at x=1

User Stobbej
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1 Answer

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By definition,


f'(x) = \displaystyle \lim_(h\to0)\frac{f(x+h)-f(x)}h

Let x = 1. Then


f'(1) = \displaystyle \lim_(h\to0)\frac{f(1+h)-f(1)}h \\\\ f'(1) = \lim_(h\to0)\frac{(3(1+h)^2+2(1+h)+1)-6}h \\\\ f'(1) = \lim_(h\to0)\frac{3+6h+3h^2+2+h+1-6}h \\\\ f'(1) = \lim_(h\to0)\frac{7h+3h^2}h \\\\ f'(1) = \lim_(h\to0)(7+3h) = \boxed{7}

User Brian Ensink
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