Calc 1 Lab 6

# Calc 1 Lab 6 - MAT 2010 Term I 10-11 LAB 6 Sept 23 FINDING...

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Unformatted text preview: MAT 2010 Term I, 10-11 LAB 6 Sept 23 FINDING SLOPES OF TANGENT LINES FOR ALL POINTS IN THE DOMAIN OF A FUNCTION AND MORE ABOUT EVALUATING LIMITS PART I FINDING A FORMULA FOR THE SLOPE OF THE TANGENT LLNE FOR ALL POINTS ON THE GRAPH OF A FUNCTION Let x be any value in the domain of the function f(x). Then we can ﬁnd the slope of the secant line though the point ( x, f(x) ) and the point ( x + h , f( x + h ) ) where h could be any positive or negative number‘but h cannot be 0 by using the formula f(-\'+h)-f(—Y) = f(xif_h)*f(x) (x+h)—x h which is of course the good old Difference Quotient. Then to ﬁnd the slope of the tangent line of the graph we compute 1~ f(X+h)—f(x) 1m h h—>0 which we say is taking the limit as h approaches 0. If this limit exists, it is the slope of the tangent line at ( x, f(x) ). The function of x we ﬁnd is called the derivative of the function f(x) and denoted f”(x). Here’s an example. Let f(x) = x2. Then f(x+h)—f(x) : (x+h)2 —x2 _2x+h h h and the limit as h approaches 0 is just 2x. So 2x is the derivative of the function x2. This also gives the slope of the tangent line for each x in the domain of the function. So for x = l, the slope of the tangent is 2; for x = 3, the slope of the tangent is 6 and for x : -4, the slope of the tangent is —8. Using this information it is easy to ﬁnd the equation of the tangent line. At the point (1,1) the slope is 2 and so the equation of the tangent line is y = 2x 1. (You should be able to ﬁnd the equation of the line.) ...
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