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4.3 What the Derivatives Tell us (part 2)

Last time:

if \( f' > 0 \) then \( f \) is increasing
if \( f' < 0 \) then \( f \) is decreasing

at \( x = c \) where \( f'(c) = 0 \) or DNE

if \( f' \) changes from \( + \) to \( - \) \( \rightarrow \) relative max at \( x = c \)
if \( f' \) changes from \( - \) to \( + \) \( \rightarrow \) relative min at \( x = c \)
no sign change across \( x = c \) \( \rightarrow \) neither max/min at \( x = c \)

Today: what does \( f'' \) tell us?

if \( f'' > 0 \), then \( (f')' > 0 \) \( \rightarrow \) \( f' \) is increasing

slope of tangent line is increasing (from left to right)

this gives us this shape:

Figure: A U-shaped curve showing f prime less than 0 on the left, f prime equals 0 at the bottom, and f prime greater than 0 on the right.

\( f'' > 0 \)

we say the graph is Concave upward

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\( f'' < 0 \), then \( (f')' < 0 \) \( \rightarrow \) \( f' \) is decreasing

moving left to right, slope of tangent line gets smaller

Figure: An inverted U-shaped curve showing f prime greater than 0 on the left, f prime equals 0 at the peak, and f prime less than 0 on the right.

\( f'' < 0 \)

we say the graph is Concave downward

concavity changes here

this is an inflection point

Figure: A curve changing from concave down to concave up at a marked inflection point.

inflection points can occur at where

\( f'' = 0 \) or \( f'' \) DNE

Similar to where critical points can occur (\( f' = 0 \) or DNE)

However, inflection pts \( \neq \) critical pts

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Example: Concavity and Inflection Points

On what intervals is \( f(x) = 3x^4 - 4x^3 \) concave up/down, and where are the inflection points?

Need \( f'' \)

\( f'(x) = 12x^3 - 12x^2 \)

\( f''(x) = 36x^2 - 24x \)

Find where \( f'' = 0 \) or DNE (very much like finding critical pts).

\( f'' = 0 \rightarrow 36x^2 - 24x = 0 \)

\( 12x(3x - 2) = 0 \rightarrow x = 0, \quad x = 2/3 \)

\( f'' \) DNE \(\rightarrow\) never because \( f(x) \) is a polynomial.

Now make a sign chart for \( f'' \).

A sign chart for f''(x) with points at x=0 and x=2/3. Signs are + for x<0, - for 0<x<2/3, and + for x>2/3. — Figure: A sign chart for f''(x) with points at x=0 and x=2/3. Signs are + for x<0, - for 02/3.

Pick any \( x < 0 \)

Check sign of \( f'' \)

\( f''(-1) > 0 \)

(concave up)

\( f''(1/2) < 0 \)

(concave down)

\( f''(1) > 0 \)

(concave up)

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Concave up: \( (-\infty, 0), (2/3, \infty) \)
Concave down: \( (0, 2/3) \)

\( f'' \) changes sign at \( x = 0 \) and \( x = 2/3 \).

So, there are two inflection pts:

\( (0, 0) \)

\( y = f(0) = 3(0)^4 - 4(0)^3 \)

\( (2/3, -16/27) \)

\( y = f(2/3) = 3(2/3)^4 - 4(2/3)^3 \)

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We can also use \( f'' \) to find relative max/min.

Graph showing a curve with a local minimum where f' = 0 and f'' > 0 , and a local maximum where f' = 0 and f'' < 0 . — Figure: Graph showing a curve with a local minimum where \( f' = 0 \) and \( f'' > 0 \), and a local maximum where \( f' = 0 \) and \( f'' < 0 \).

relative max: \( f' = 0 \), \( f'' < 0 \)

relative min: \( f' = 0 \), \( f'' > 0 \)

Second Derivative Test (for finding relative max/min)

If \( x = c \) is a critical number (\( f' = 0 \) or \( f' \text{ DNE} \)) and if:

\( f''(c) > 0 \) then there is a relative min at \( x = c \)
\( f''(c) < 0 \) then there is a relative max at \( x = c \)
\( f''(c) = 0 \) then this test is inconclusive

(\( x = c \) could still be a relative max/min)

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Example

\( f(x) = e^x(x - 1) \)
Again, need \( f'' \)

Product Rule:

\( f'(x) = e^x(1) + (x - 1)e^x = e^x + xe^x - e^x = xe^x \)

\( f''(x) = xe^x + e^x(1) = xe^x + e^x = e^x(x + 1) \)

\( f'' = 0 \rightarrow e^x = 0 \) or \( x + 1 = 0 \)

\( ↓ \)

never, since \( e^x \neq 0 \)

\( x = -1 \)

\( f'' \text{ DNE} \rightarrow \text{never, since } e^x \text{ and } x + 1 \text{ exist for all } x \)

Sign chart for f'' with a critical point at x = -1 , showing negative signs and concave down shape to the left, and positive signs and concave up shape to the right. — Figure: Sign chart for \( f'' \) with a critical point at \( x = -1 \), showing negative signs and concave down shape to the left, and positive signs and concave up shape to the right.

\( f''(-2) = e^{-2}(-2 + 1) < 0 \)

Concave up: \( (-1, ∞) \)

Concave down: \( (-∞, -1) \)

\( f'' \) changes sign at \( x = -1 \)
so inflection pt @ \( (-1, -2e^{-1}) \)

\( f(-1) = e^{-1}(-1 - 1) \)

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Example: Second Derivative Test

Consider the function from the previous example:

\[ f(x) = e^x(x-1) \]

Process to find max/min: find critical numbers first.

\[ f'(x) = xe^x = 0 \rightarrow \text{notebox: } x=0 \rightarrow \text{the only critical #} \]

\( f'(x) = xe^x \) DNE doesn't happen.

Checking Concavity

Now check sign of \( f'' \) at each critical number.

\[ f''(x) = e^x(x+1) \]\[ f''(0) = e^0(0+1) > 0 \]

So, near this critical # \( x=0 \) the graph is concave up.

So there is a relative minimum at \( x=0 \).

Figure: A sketch of a concave up parabola-like curve with a horizontal tangent line at its minimum point.

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Example: Polynomial Extrema

\[ f(x) = x^4 - 4x^3 + 1 \]\[ f'(x) = 4x^3 - 12x^2 \]

Find critical pts:

\[ f' = 0 \rightarrow 4x^3 - 12x^2 = 0 \]\[ 4x^2(x-3) = 0 \]\[ \text{notebox: } x=0, x=3 \quad \text{two critical pts} \]

Second Derivative Test

\[ f''(x) = 12x^2 - 24x \]\[ f''(3) = 36 > 0 \rightarrow \text{notebox: relative min at } x=3 \]\[ f''(0) = 0 \rightarrow \text{inconclusive} \]

We still need to find out what's going on. Use first derivative test as backup.

First Derivative Test Analysis

Figure: A sign chart for f prime on a number line with critical points at x=0 and x=3. Signs are negative for both intervals x < 0 and 0 < x < 3.

No sign change across \( x=0 \), so there is neither a rel. max nor a rel. min at \( x=0 \).

\( f'(-1) < 0 \)

\( f'(1) < 0 \)