U5: Analytical Applications of Differentiation

Mean Value Theorem

If f(x) is a function that is continuous on the closed intervals [a,b] and differentiable on the open interval (a,b), then there must exist a value c between (a,b)

f'(c) = (f(b) - f(a))/(b - a)

● Example: Confirm f(x) = x^3 on [0, 3] and find a value that satisfies this theorem

○ f’(x)=3x2

○ 3x2 = (f(3) − f(0))/(3−0) → 3x2 = ((3)3 − ((0)3)/3 → 3x2 = 9 → x ± √3

Function Increasing or Decreasing

● f(x) is increasing when f’(x) is positive

● f(x) is decreasing when f’(x) is negative

Extreme Value Theorem

If f(x) is continuous on a closed interval [a,b], then f(x) has both a minimum and maximum on the interval.

Example: Locate the absolute extrema on the function on the closed interval.

● f(x) = x3 - 12x [0, 4]

● f’(x) = 3x2 - 12

● 0 = 3x2 - 12

● x = -2, 2

● f(0) = 0, f(2) = -16, f(4) = 16

○ f(x) has an absolute minimum at x = 2 and an absolute maximum at x = 4

First Derivative Test

For first derivative tests, derive the function once and set it to 0. After that, find the zeros and plug them into a number line. Using your derived function, plug-in numbers before and after your constant (the zeros of the function) to see if it becomes negative or positive, as shown below.

● If it’s positive, constant, negative then it’s a relative maximum

● If it’s negative, constant, positive then it’s a relative minimum

● f(x) has a relative maximum at x = c

● f(x) has a relative minimum at x = c

Example:

● f(x) = x2 + 6x + 10

● f’(x) = 2x + 6

● 0 = 2x + 6

● x = -3

● f(x) is decreasing (-∞,-3)

● f(x) is increasing at (-3, +∞)

● f(x) has a relative minimum at x = -3

Concavity

● The graph of f is concave up when f’(x) is increasing

● The graph of f is concave down when f’(x) is decreasing

● If f’’(x) is positive then the graph of f is concave up

● If f’’(x) is negative then the graph of f is concave down

● Points of inflection

○ Occurs when f(x) changes concavity

○ Determined by a sign change for f’’(x)

Second Derivative Test

Example:

● y = -x3 + 3x2 - 2

● y’ = -3x2 + 6x

● y’’ = -6x + 6

● 0 = -6x + 6

○ x = 1

● Like the first derivative test, after you find your second derivative, find the zeros of f(x).

● In this case, 1 is the zero, so it goes on the number line.

● To see if it’s concave up or concave down, use -1 and 2 to plug into the second derivative function.

● When you plug in -1 you get a positive number.

● When you plug in 2 you get a negative number.

f(x) is concave up on (-∞,1) because f’’(x) is positive on that interval.

f(x) is concave down on (1, +∞) because f’’(x) is negative on that interval.

Critical Numbers

● Critical numbers are points on the graph of a function where there’s a change in direction.

● To find critical numbers, you use the first derivative of the function and set it to zero.

Example:

● f(x) = 2secx + tanx

● f’(x) = 2secxtanx + sec2x

● 0 = secx(2tanx + secx) → 0 = secx → 0 = 1/cosx → x = 3π/2, π/2

● 0 = 2tanx + secx → -secx = 2tanx → -1/cosx = 2sinx/cosx → -1 = 2sinx → sinx = -½ → x = 7π/6, 11π/6

● The critical numbers for this function are x = 3π/2, π/2, 7π/6, 11π/6