To graph functions in calculus we neeed to review several theorem. 3 theorems have been used to find maxima and minima using first and second derivatives and they will be used to graph functions. We need 2 more theorems to be able to graphs functions using first and second derivatives.

Theorem 4: If f is differentiable on an interval (I1 , I2) and differentiable on [I1 , I2] and

4.a - If f ' (x) > 0 on (I1 , I2) , then f is increasing on [I1 , I2].

4.b - If f ' (x) < 0 on (I1 , I2) , then f is decreasing on [I1 , I2].

4.c - If f ' (x) = 0 on (I1 , I2) , then f is constant on [I1 , I2].

Theorem 5: If f is twice differentiable on an interval (I1 , I2) and

5.a - If f '' (x) > 0 on (I1 , I2) , then f has concavity up on [I1 , I2].

5.b - If f ' (x) < 0 on (I1 , I2) , then f is concavity down [I1 , I2].

We will present examples of graphing functions using the theorems in "using first and second derivatives" and theorems 4 and 5 above.

__Example 1:__ Use first and second derivative theorems to graph function f defined by

f(x) = x^{ 2}

__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: f ' (x) = 2x = 0 , which gives x = 0.

The sign of f ' (x) is given in the table below. f ' (x) is negative on (-infinity , 0) f decreases on this interval. f ' (x) is positive on (0 , +infinity) f increases on this interval. Also according to theorem 2(part a) "using first and second derivatives", f has a minimum at x = 0.

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 (-infinity , +infinity) 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 (+infinity) or vey small (-infinity) , f(x) = x^{ 2} becomes very large.

See table above and graph below.

__Example 2:__ Use first and second derivative theorems to graph function f defined by

f(x) = x^{ 3} - 4x^{ 2} + 4x

__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 (+infinity) , f(x) = x^{ 3} - 4x^{ 2} + 4x becomes very large (+infinity). As x becomes very small (-infinity), f(x) becomes very small (-infinity).

step 4: The table and the graph are shown below.

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calculus problems