if f(1)=2 and f'(1)=5, use the equation of the line tangent to the graph of f at x=1 to approximate f(1.2).
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Originally Posted by asnxbbyx113 if f(1)=2 and f'(1)=5, use the equation of the line tangent to the graph of f at x=1 to approximate f(1.2). Point-Slope Form? f(1) = 2 gives the point f'(1) = 5 gives the slope. y - 2 = 5(x-1) Substitute x = 1.2 That's about it.
Originally Posted by asnxbbyx113 if f(1)=2 and f'(1)=5, use the equation of the line tangent to the graph of f at x=1 to approximate f(1.2). Or you can think of it as a truncated Taylor series. $\displaystyle f(1.2) \approx f(1) + \frac{1}{1!}f^{\prime}(1) \cdot 0.2$ So $\displaystyle f(1.2) \approx 2 + 5 \cdot 0.2$ etc. -Dan
Originally Posted by topsquark Or you can think of it as a truncated Taylor series. $\displaystyle f(1.2) \approx f(1) + \frac{1}{1!}f^{\prime}(1) \cdot 0.2$ So $\displaystyle f(1.2) \approx 2 + 5 \cdot 0.2$ etc. -Dan That is not the best explanation to give this poster. Because he will not hear the term "Taylor Series" for another year.
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