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# Trajectory high point A stone is launched vertically ISBN: 9780321570567 2

## Solution for problem 43E Chapter 4.1

Calculus: Early Transcendentals | 1st Edition

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Problem 43E

Trajectory high point A stone is launched vertically upward from a cliff 192 ft above the ground at a speed of 64 ft/s. Its height above the ground ?t seconds after the launch is g?iven by? ?s =?? 6?t2+ 64?t + 192? for ? ? 6. When does the stone reach its maximum height?

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STEP_BY_STEP SOLUTION Step-1 Let f be a continuous function defined on an open interval containing a number ‘c’.The number ‘c’ is critical value ( or critical number ). If f (c) 1 1 = 0 or f (c) is undefined. A critical point on that graph of f has the form (c,f(c)). Step-2 When an output value of a function is a maximum or a minimum over the entire domain of the function, the value is called the absolute maximum or the absolute minimum. Let f be a function with domain D and let c be a fixed constant in D . Then the output value f ) is the 1. Absolute maximum value of f on D if and only if f(x) f(c) , for all x in D. 2. Absolute minimum value of f on D if and only if f(c) f(x) , for all x in D. Step_3 Given is; A stone is launched vertically upward from the cliff 192 ft above the ground at a speed of 64 ft/s , and also given that its height above the ground t seconds after the launch is given by 2 S = -16t +64t+192 for 0 t 6. Now , we have to find out the value of stone reach its maximum height . For that we have to evaluate the function at critical points.so, the critical points satisfies the equation . 2 Given function is; S = -16t +64t+192 . For the critical points we have to differentiate the function with respect to t .[ Since from the step-1] d d 2 dt(s) = dt( -16t +64t+192 ) s = d ( -16t ) + d...

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##### ISBN: 9780321570567

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