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Unformatted text preview: AMS 151 (Fall, 2009) Joe Mitchell Applied Calculus I Practice Problems for Quiz # 4 – Solution Notes 1. Find the equation of the tangent line to the curve y = 4 x 3 + 1 at the point (2,33). Since the slope is given by the derivative function, y ′ ( x ) = 4 · 3 x 2 = 12 x 2 , we know that the slope of the tangent line at (2,33) is y ′ (2) = 48. Thus, the equation of the tangent line at (2,33) has the form y = 48 x + b , and we can find b using the fact that the line must pass through the point (2,33): 33 = 48 · 2 + b , implying that b = 63. Thus, the equation of the tangent line at (2,33) is y = 48 x 63. 2. Find the equation of the tangent line to the curve y = 2 x 2 5 x + 3 at the point where x = 3. Since the slope is given by the derivative function, y ′ ( x ) = 2 · 2 x 5 = 4 x 5, we know that the slope of the tangent line at the point where x = 3 (namely, the point (3 , 2 · 3 2 5 · 3 + 3) = (3 , 6) is y ′ (3) = 7. Thus, the equation of the tangent line at (3,6) has the form y = 7 x + b , and we can find b using the fact that the line must pass through the point (3,6): 6 = 7 · 3 + b , implying that b = 15. Thus, the equation of the tangent line at (3,6) is y = 7 x 15. 3. Evaluate the limit lim x →∞ 5 x 3 2 x 2 + 7 2 x 4 x 2 7 x 3 5 x 3 2 x 2 + 7 2 x 4 x 2 7 x 3 = 5 x 3 2 x 2 + 7 2 x 4 x 2 7 x 3 · 1 /x 3 1 /x 3 = 5 2 x + 7 x 3 2 x 2 4 x 7 Now, as x goes to infinity, we see that the limit is 5 / ( 7), since all other terms with x in the denominator vanish...
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This note was uploaded on 02/07/2010 for the course AMS 151 taught by Professor Zhang during the Fall '08 term at SUNY Stony Brook.
 Fall '08
 Zhang

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