f(x) and f’(x) from the beginning Say I want to draw the graph of f (X) = 4X – X 2 and include some information about its ”lutning” at certain points. Draw between X = -1 and 5 f (X) = 4X – X 2 f’(X) = 4 – 2X Xf(X)f’(X)=4-2XPunkt -56(-1,-5) 004(0,0) 132(1,3) 240(2,4) 33-2(3,3) 40-4(4,0) 5-5-6(5,-5)

f(x) and f’(x) from the beginning NOTE f(X) gives the Y coordinate. f’(x) gives the LUTNING of the curve at that point. For example when x = 1, at coordinate (1,3), the gradient is 2. THIS IS THE GRADIENT OF THE TANGENT TO THE CURVE. THE TANGENT IS A STRAIGHT LINE WHICH TOUCHS THE CURVE AT POINT (1,3) ONLY. The tangent has an equation of Y=KX + M. Point (1,3) is ”shared” between the curve and the line (tangent). Xf(X)f’(X)=4-2XPunkt -56(-1,5) 004(0,0) 132(1,3) 240(2,4) 33-2(3,3) 40-4(4,0) 5-5-6(5,-5)

f(x) and f’(x) from the beginning The graph looks like this. I have drawn a tangent to the point (2,4) where f’(X) = 0. That is easy and the equation of the line is Y=4 because K=0. How would I find the equation of the tangent to the point (1,3)? For example, I often do this to show you examples of tangents to curves because I am terrible at drawing. Ask Shuzo and Niels! Well, anybody really!

f(x) and f’(x) from the beginning At (1,3), X=1, Y= 3 and K = 2 because f’(X) = 2. Y= KX + M 3 = M  M=1. Y= 2X + 1 Xf(X)f’(X)=4-2XPunkt -56(-1,5) 004(0,0) 132(1,3) 240(2,4) 33-2(3,3) 40-4(4,0) 5-5-6(5,-5)

f(x) and f’(x) from the beginning Here is the tangent to the curve.

f(x) and f’(x) from the beginning At (3,3), X=3, Y= 3 and K = -2 because f’(X) = -2. Y= KX + M 3 = -2*3 + M  M=9. Y= -2X + 9 Xf(X)f’(X)=4-2XPunkt -56(-1,5) 004(0,0) 132(1,3) 240(2,4) 33-2(3,3) 40-4(4,0) 5-5-6(5,-5)

f(x) and f’(x) from the beginning Tangent to (3,3) has equation Y= -2X + 9

Exam Example För funktionerna f och g gäller att f (x) = 5x 2 + 3x och g(x) = x 2 +8x a) Bestäm det värde på x där grafen till f har lutningen 18 b) Grafen till g har en tangent i den punkt där x = 6 Bestäm koordinaterna för tangentens skärningspunkt med x-axeln. a)This means finding the value of X when f’(x)= 18. f (x) = 5x 2 + 3x  f’(x) = 10X+3.  10X + 3 = 18. Answer: X=1,5 b)See next slide.

Exam Example b) g(x) = x 2 +8x. On this curve, when x= 6, g(x) = = 84. REMEMBER THAT THE TANGENT “SHARES” THIS POINT WITH THE CURVE therefore the tangent goes through the coordinate (6,84). To find K for g(x) = x 2 +8x use g’(x) = 2x+8. When x = 6, g’(x) = 20. Therefore the tangent to the curve has k = 20. The tangent goes through the coordinate (6,84). X=6, Y=84, K= = m  m= -36 Equation of the straight line: Y = 20X – 36 Bestäm koordinaterna för tangentens skärningspunkt med x-axeln. OBS!! NOT the Y axis! This is when y= 0  0 = 20x – 36  x= 1,8. Answer: (1,8;0)

Summary f’(2) below = 0. A tangent has been drawn at that point. f(x) = 4x – x 2 f’(x) = 4 -2x If we choose x=3, then f’(3) = - 2. This is also the gradient of the tangent at the point (3,3)

Summary And here it is.

Summary Imagine you are in an exam. You are asked to find the equation of, for example, the tangent to f(x) = x 3 where x=2. If you were to panic, you might forget some of this f(x) and f’(x) stuff! What is important is to remind yourself of what f(x) and f’(x) really MEAN. JUST IMAGINE YOU WERE GOING TO DRAW THE GRAPH THEN ATTEMPT TO DRAW THE TANGENT AT THAT POINT. That way you could get an idea of the equation of the line couldn’t you? Of course, to draw the graph you WOULD USE f(x) and NOT f’(x). So when X=2, the point on the graph you are drawing would have a y coordinate of f(x) = 8. So you would know that the coordinate (2, 8) is on the graph. The LUTNING at that coordinate is given by f’(x) = 3x 2. = 12. THEREFORE YOU WOULD ATTEMPT TO DRAW A TANGENT TOUCHING JUST (2,8) with k = 12. We now know x, y and k for the tangent. We only need to find M for Y=KX+M. X=2; Y= 8; K= 12 therefore M=y-kx = M = -16 so the equation of the tangent to y=x 3 where x= 2 is Y=12X - 16

Finally…. This graph is an ”x squared” graph. Note the following facts:- The graph is symmetrical as you can see. The gradient when x=0 is 4 and when x = 4, the gradient is -4. The same situation is true for x=1 and x=3. The highest value of f(x) happens when f’(x) = 0. This is not a coincidence as we will see later.