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Chap3_Sec1

# Chap3_Sec1 - of f at 0 that is So f’ x = f’(0 a x 1 lim...

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Compute the derivative of f(x) = x 8 + 12 x 5 – 4 x 4 + 10 x 3 – 6 x + 5 NEW DERIVATIVES FROM OLD Example 5 (3.1)

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Find the points on the curve y = x 4 - 6 x 2 + 4 where the tangent line is horizontal. NEW DERIVATIVES FROM OLD Example 6 (3.1)
So, the given curve has horizontal tangents when x = 0, , and - . s The corresponding points are (0, 4), ( , -5), and (- , -5). NEW DERIVATIVES FROM OLD Example 6 3 3 3 3

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The equation of motion of a particle is s = 2 t 3 - 5 t 2 + 3 t + 4, where s is measured in centimeters and t in seconds. s Find the acceleration as a function of time. s What is the acceleration after 2 seconds? NEW DERIVATIVES FROM OLD Example 7 (3.1)
Let’s try to compute the derivative of the exponential function f ( x ) = a x using the definition of a derivative: 0 0 0 0 ( ) ( ) '( ) lim lim ( 1) lim lim x h x h h x h x x h h h f x h f x a a f x h h a a a a a h h + + - - = = - - = = EXPONENTIAL FUNCTIONS

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The factor a x doesn’t depend on h . So, we can take it in front of the limit: s Notice that the limit is the value of the derivative

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Unformatted text preview: of f at 0, that is, So f’ ( x ) = f’ (0) a x 1 '( ) lim h x h a f x a h →-= 1 lim '(0) h h a f h →-= EXPONENTIAL FUNCTIONS If f ( x ) = e x -x, find f’ and f’’ . Compare the graphs of f and f’ . EXPONENTIAL FUNCTIONS Example 8 (3.1) The function f and its derivative f’ are graphed here. s Notice that f has a horizontal tangent when x = 0. s This corresponds to the fact that f’ (0) = 0 . EXPONENTIAL FUNCTIONS Example 8 Notice also that, for x > , f’ ( x ) is positive and f is increasing. When x < , f’ ( x ) is negative and f is decreasing. At what point on the curve y = e x is the tangent line parallel to the line y = 2 x ? EXPONENTIAL FUNCTIONS Example 9 (3.1) Thus, the required point is: ( a , e a ) = ( l n 2, 2) EXPONENTIAL FUNCTIONS Example 9...
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Chap3_Sec1 - of f at 0 that is So f’ x = f’(0 a x 1 lim...

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