A tutorial on how to use the first and second derivatives, in calculus, to study the properties of the graphs of functions.
Example 1Use first and second derivative theorems to graph function f defined by
Solution to Example 1.
step 1: Find the first derivative, any stationary points and the sign of f ' (x) to find intervals where f increases or decreases.
f ' (x) = 2x
The stationary points are solutions to:
step 2: Find the second derivative, its signs and any information about concavity.
f ''(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.
step 3: Find any x and y intercepts and extrema.
y intercept = f(0) = 0.
x intercepts are found by solving f(x) = x 2 = 0. x intercept = 0.
From the signs of f ' and f'', there is a minimum at x = 0 which gives the minimum point at (0 , 0).
step 4: Put all information in a table and graph f.
Also as x becomes very large (+?) or veyy small (-?) , f(x) = x 2 becomes very large.
See table above and graph below.
Example 2Use first and second derivative theorems to graph function f defined by
Solution to Example 2.
step 1: f ' (x) = 3x 2 - 8x + 4.
Solve 3x 2 - 8x + 4 = 0
solutions are: x = 2 and x = 2/3, see table of sign below that also shows interval of increase/decrease and maximum and minimum points.
step 2: f '' (x) = 6x - 8.
Solve 6x - 8 = 0 ; solution is x = 4/3 inflection point: where concavity changes. See sign and concavity in table below.
step 3: y intercept = f(0) = 0. x intercepts solve x 3 - 4x 2 + 4x = 0.
factor x out x(x 2 - 4x + 4) = 0 and solve the quadratic equation x 2 - 4x + 4 = (x - 2) 2 = 0 to find solutions: x = 0, x = 2 of multiplicity 2.
Also as x becomes very large (+?) , f(x) = x 3 - 4x 2 + 4x becomes very large (+?). As x becomes very small (-?), f(x) becomes very small (-?).
step 4: The table and the graph are shown below.
More references on calculus problems