S5 Calculus Tangents and Stationary Points
By Jay Teach
S5 Calculus Tangents and Stationary Points
·# Assess (using https://www.menti.com/5fteqn32yx ) their understanding of gradients by pairing equations of straight lines with their graphs, (graphs available at https://www.desmos.com/calculator/pmqizkrhxq ): ·●y=2(x+3.5) ●y=-1/4(5x-24) ●y=4(x+1)-1 ●y=-1/2x-2 · ·# Assess self-reported confidence in: [If not confident, do them.] ·Differentiating ●x⁻², ●1/(2x), ●3x²-5x+1, ●3/sqrt(x³), ●(x+2)(2x-1) · · ·# Recap: Basic Differentiation Rules ·d/dx( f(x) ) = f’(x) = df/dx The derivative of a function is the instantaneous rate of change of that function at each point. · ·d/dx( x^n ) = nx^(n-1) To differentiate: by the old power, decrease the power by one. · ·d/dx( a ) = 0 The derivative of a constant is zero. · ·d/dx( ax^n ) = a d/dx(x^n) = a nx^(n-1) When differentiating a constant times a power of x, extract the constant and multiply it from the outside. · ·d/dx( f(x)+g(x) ) = f’(x)+g’(x) The derivative of a sum of functions is the sum of their derivatives. · · ·# Recap: Finding the equation of a line between two points ·●Find m = Δy/Δx ●Put this into y=mx+c ●Put a coordinate-pair into the equation to solve for c ●Re-write y=mx+c with values for m and c ·Example Ex.: Find the equation of the line passing through (2,4) , (3,9) · · ·# Differentiating from First Principles ·We construct a crude tangent line between two points on f(x) ·The first point has coordinates (a, f(a) ) ·The second point is shifted-over by ‘h’, with coordinates (a+h, f(a+h) ) ·●y = mx+c ●m = (f(a+h)-f(a))/((a+h)-(a)) ●c = f(a)-ma · ·# What is the pattern as we zoom in and make ‘h’ shrink? (Discuss in pairs) ·Experiment with some real numbers: a=2 and h=1, decreasing ‘h’ to 0.5, 0.25, 0.1 etc. ·Conclusion: shrinking h gives a better-and-better fit to the f(x) graph. ·[Differentiation: canned answer for potential hard question: We aren’t ‘dividing by zero’ because 1. h cancels-out from the top and bottom, and 2. we never actually go to h=0, we just go near it → problem dodged 👌! · ·# The slope of the tangent line (m) at any point on f(x) EQUALS the value of f’(x) for that point ·This expression for ‘m’ is the DEFINITION of ‘Differentiating’. [Write eq] ·(If you’re keen, ask me after for the proof for how it works for x^n) · · ·# We can now use f’(x) to gain knowledge about changes in f(x) ·Task: sketch f(x)=x² , on another pair of axes below it sketch f’(x), below this sketch a table horizontally of f’(-2),f’(-1),f’(0),f’(1),f’(2), evaluate f’(x) at those points. ·If the result is positive/negative/zero, below it write +/-/0 ·What’s the pattern (Discuss): When f’(x) is -, f(x) is decreasing, when f’(x) is +, f(x) is increasing, when f’(x) is 0, f(x) is stationary. · ·# Exercise: For f(x)=x²-4x, 1. find f’(x) 2. Sketch the graph of f(x) 3. When does f’(x)=0 ? 4. Sketch the graph of f’(x) (solutions: https://ssddproblems.com/differentiate-y-fx/ ) · ·# Every value of ‘a’ that makes f’(a)=0 is called a “stationary point” ·Does the stationary point always line up with the minimum/maximum of the parabola? Re-write these parabolas in point-intercept form to investigate! ·In the form y=a(x+b)²+h ●f(x)=x²+4x+4 ●p(x)=-x²-8x ●s(x)=x²+6x-3 ·After doing it by hand, you can check your results by typing them into today’s Desmos graph. · ·# Extra Exercises: page 102 Ex. 2,3,4 Page103-104 Examples16–18, 6L2 · ·Plenary: Assess self-reported confidence in: ·●Pairing up a particular graph with its derivative graph. ●Identifying where f’(x) is +,-,0 just by looking at f(x). ●Sketching f'(x) just by looking at f(x). ●Using f(x) to explain what happens in f’(x). ●Using f'(x) to explain what happens in f(x). ●Using the equation for f'(x) to calculate the rate of change of f(x) at different places.



