1,3. f x x f x x. Lim. Lim. Lim. Lim Lim. y 13x b b 10 b So the equation of the tangent line is y 13x

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1 1.5 Topics: The Derivative lutions 1. Use the limit definition of derivative (the one with x in it) to find f x given f x 4x 5x 6 4 x x 5 x x 6 4x 5x 6 f x x f x f x x0 x x0 x xx x x x x x = x 0 x = = x 4x 8xx 4 x 5x 5x 6 4x 5x 6 x0 x 4x x 0 8xx 4 x 5x 5x 6 8xx 4 x 5x = x 0 x x 8x 4x 5 x = x0 x x0 x 0 4x 8x 4x 5 x = 8x 4x 5 8x x 5 x 5x 6 a. Use your formula from problem 1 to find the equation of the tangent line to 1,3 f x 8x 5 so at 1,3 f y 13x b b 10 b the equation of the tangent line is y 13x 10 b. Use the alternate limit definition of derivative to find f 1 given answer at x = 1 with your formula from problem 1.? 1 4x 5x 6 3 4x 5x f x 4x 5x 6 at the point f x 4x 5x 6. Did you get the same f 1 4x x1 x 1 x1 x 1 x1 x 1 x1 x 1 x1 YES, I got the same answer for the slope of the tangent line at x = 1 using this alternate formula as when I used the general derivative formula in problem #1 and substituted in x = Use your formula from problem 1 to help you find out where,, xy, f x 4x 5x 6 has a horizontal tangent line. If we want to know WHERE the original function has a horizontal tangent line then we will just set our formula for the derivative equal to zero (horizontal lines have a slope of zero) and solve for x.

2 x 5 0 x f y our function will have a horizontal tangent line AT, Find the equation of the two tangent lines to the graph of f x4x x through the point,5 f x 4x x f x x0 x0 4x x0 4 x x x x 4x x f x x f x x x 4xx xx x 4x x x x x0 4 xx x 4 x 0 4 x Note: Rough Sketch Let ( a, b ) be a point of tangency in the diagram. The slope of the line at ( a, b ) will be 4 a since the derivative formula is 4 x. Since the tangent line passing through ( a, b ) also passes through (, 5 ) the slope formula tells us that the slope will be b 5 a. b 5 4 a b 5 a 4 a b 5 a 8a 8 a 8a b 3 0. Since the point ( a, b ) lies on a f x 4x x we know that b 4a a. the graph of Substituting into gives us a 8a 4a a 3 0 a 4a 3 0 a 1, a 3 a 8a b 3 0 the two points of tangency (and corresponding slope) are tangent lines are y x 1 and y x 9 5. For each of the given graphs y f x sketch the graph of y f x 1,3 slope and 3,3 slope. the equations of the. A) y f x y f x

3 B) y f x y f x C) y f x y f x 6. The following graph is a picture of gx (the derivative of g x ) y g x Find the following g 0 ph!) A) 3 (this is the "y" value when x is 0...look at the gra B) on what interval(s) is gx increasing? C) on what interval(s) is gx decreasing? gx will increase when gx will decrease when D) Where (x-value) will gx have a relative maximum? This will occur when E) Where (x-value) will gx have a relative minimum? This will occur when g x is positive so, 3 and 3, x is negative so 3,3 g x g x g changes from positive to negative so x 3 changes from negative to positive so x 3

4 7. f x x 1 Find the derivative from the left and right at x = 1. f x differentiable at x = 1? x 1 if x 1 0 x 1 if x 1 f x x 1 x 1 if x 1 0 x 1 if x 1 the derivative from the left will be the derivative of f x x 1 and the derivative from the right will be the derivative of f x x 1. f x x f xx x1x 1 x x 1 x 1 x f x x 1 f x 1 1 xx xx xx xx x f x x f xx x1x 1 x x 1 x 1 x f x x 1 f x 1 1 xx xx xx xx x 8. f x is NOT differentiable at x = 1 because the derivative from the left does not equal the derivative from the right. x x 1 g x 4x3 x gx continuous at x =? Yes, gx is continuous at x = because if you substitute x = into both pieces of the function you get the same y value of 5. This means the two function pieces match up at x = so the function is continuous there. gx differentiable at x =? To answer this question we must find the derivative of each piece of the function and evaluate each of those pieces at x =. The derivative from the left at x = will be the derivative of xx x x x x g x x 1 g x x x x The derivative from the right at x = will be the derivative of g x 4x 3 evaluated at x = x x x x x g x 4x 3 g x x x x x g x x 1evaluated at x =. x x x gx IS differentiable at x = because the derivative from the left equals the derivative from the right!

5 9. hx 3x x1 x 1 x 1 hx continuous at x = 1? Yes, hx is continuous at x = 1 because if you substitute x = 1 into both pieces of the function you get the same y value of 1. This means the two function pieces match up at x = 1 so the function is continuous there. hx differentiable at x = 1? To answer this question we must find the derivative of each piece of the function and evaluate each of those pieces at x = 1. The derivative from the left at x = 1 will be the derivative of x x x x 1 h x x 1 h x x1 x1 x1 x1 1 1 x1 x1 h x h x x x 1 x 1 x 1 x 1 x The derivative from the right at x = 1 will be the derivative of hx 3x evaluated at x = x 1 x 1 x 1 x 1 x 1 h x h x x h x 3x h x x1 x1 x1 x1 h x x 1evaluated at x = 1. x1 3 3 hx is NOT differentiable at x = 1 since the derivative form the left does not equal the derivative from the right.