Now each term of the denominator contains an h but all contain higher powers of h than 1 except for the first term, therefore when we divide each term by h and take the limit as h → 0, all terms but the first vanish, leaving Here is the derivative expression for this function it's just a matter of plugging the function into the difference quotient expression: The function (see graph below right) has a positive slope for all x in its domain, except for x = 0, at which the slope is undefined. It has a vertical asymptote at x = 1, therefore finding the slope of f(x) at x = 1 makes no sense – the function has no actual value there. This is an interesting function–a rational function. We'll wait for a while to actually prove that this limit is the slope of f(x) at x. The graphs below will help you visualize how making the distance between our two points smaller gives us a better and better estimate of the slope of f(x) at x. That is the essence of differential calculus: Finding the limit of a slope function as the change in the independent variable approaches zero. ![]() Now do the following thought exercise: Imagine that Δx gets smaller and smaller, and eventually vanishes or becomes "infinitessimally small." In the limit that Δx = 0, we would have the exact slope of our function at point x. Convince yourself that this equation still just represents rise over run, Δy / Δx (and don't forget that y = f(x)). The slope of our secant line ( magenta), m, is written above the graph in terms of x, Δx, f(x) and f(x+Δx). This will allow us to reduce the width of the interval between them (the x-distance) by reducing Δx. To do that, we'll first change our notation a little and use x and ( x + Δx) instead of x 1 and x 2 to label our two points. Step 4: Put all information in a table and graph f.This time we'll try to find the slope of our curve at some specific point x. x intercept = 0.įrom the signs of f ' and f'', there is a minimum at x = 0 which gives the minimum point at (0, 0). X intercepts are found by solving f(x) = x 2 = 0. Step 3: Find any x and y intercepts and extrema. Step 2: Find the second derivative, its signs and any information about concavity.į ''(x) = 2 and is always positive (this confirms the fact that f has a minimum value at x = 0 since f ''(0) = 2, theorem 3(part a)), the graph of f will be concave up on (-∞, +∞) according to theorem 5(part a) above. Also according to theorem 2(part a) "using first and second derivatives", f has a minimum at x = 0. f ' (x) is positive on (0, ∞) f increases on this interval. f ' (x) is negative on (-∞, 0) f decreases on this interval. ![]() The sign of f ' (x) is given in the table below. Step 1: Find the first derivative, any stationary points and the sign of f ' (x) to find intervals where f increases or decreases. Use first and second derivative theorems to graph function f defined by We will present examples of graphing functions using the theorems in "using first and second derivatives" and theorems 4 and 5 above. ĥ.b - If f ' (x) < 0 on (I1, I2), then f is concavity down. Ĥ.b - If f ' (x) 0 on (I1, I2), then f has concavity up on. Theorem 4: If f is differentiable on an interval (I1, I2) and differentiable on andĤ.a - If f ' (x) > 0 on (I1, I2), then f is increasing on. We need 2 more theorems to be able to study the graphs of functions using first and second derivatives. ![]() 3 theorems have been used to find maxima and minima using first and second derivatives and they will be used to graph functions. ![]() To graph functions in calculus we first review several theorem. First, Second Derivatives and Graphs of FunctionsĪ tutorial on how to use the first and second derivatives, in calculus, to study the properties of the graphs of functions.
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