f(x) = e^(sin2x) , 0 ≤ x ≤ pi (a). Use calculus to find the coordinates of the turning points on the graph of y = f(x)
First, understand the concept of a turning point (zero gradient, implying first derivative is zero at the turning point)How do we differentiate exponentials ( if f(x) = e^g(x) then f'(x) = g'(x)*e^g(x) ) Okay, now how do we differentiate (sin2x) (=g(x) in this example)Chain rule & recap how to differentiate trig (we require familiarity with d(sinx)/dx = cosx)Figure out f'(x) and then set to equal zeroSolve equation to find coordinates of turning pointsOf course, use whiteboard and step by step lines of working to guide to answer
MW
Related Maths A Level answers
All answers ▸Solve the differential equation dx/dt = -2(x-6)^(1/2) for t in terms of x given that x = 70 when t = 0.
Find the area under the curve with equation y = 5x - 2x^2 - 2, bounded by the x-axis and the points at which the curve reach the x-axis.
The function f has domain (-∞, 0) and is defines as f(x) = (x^2 + 2)/(x^2 + 5) (here ^ is used to represent a power). Show that f'(x) < 0. What is the range of f?
The points A and B have coordinates (3, 4) and (7, 6) respectively. The straight line l passes through A and is perpendicular to AB. Find an equation for l, giving your answer in the form ax + by + c = 0, where a, b and c are integers.