A curve is defined for x > 0. The gradient of the curve at the point (x,y) is given by dy/dx = x^(3/2)-2x. Show that this curve has a minimum point and find it.
This is a typical exam style question, taken from an AQA paper. This question is testing your knowledge of stationary points and differentiation. Step 1: Find all stationary points by setting the first derivate to 0, and solving the equation. Step 2: Determine what type of stationary points those we found in step 1 are. This is done by obtaining the second derivative, and substituting in the x values found in step 1. (Optional step 3: interpretationFirst derivative - gradientSecond derivative - rate of change of gradient)
YC
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For a curve of gradient dy/dx = (2/(x^2))-x/4, determine a) d^2y/dx^2 b) the stationary point where y=5/2 c) whether this is a maximum or minmum point and d) the equation of the curve
The function f is defined for all real values of x as f(x) = c + 8x - x^2, where c is a constant. Given that the range of f is f(x) <= 19, find the value of c. Given instead that ff(2) = 8, find the possible values of c.
What is the natural logarithm?