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4.2 The Mean Value Theorem

Rolle's Theorem

If \( f(x) \) is:

continuous on \( [a, b] \) → no holes, no asymptotes, no gaps
differentiable on \( (a, b) \) → smooth, no corners, no cusps
\( f(a) = f(b) \) → same y value at ends of interval

then there is at least one value of \( x \), call it \( c \), \( a < c < b \) where \( f'(c) = 0 \) → at least one place between ends where tangent line is horizontal.

Figure: Graph of a smooth curve starting and ending at the same y-value with a horizontal tangent at peak c.

Smooth curve satisfying Rolle's Theorem.

Figure: Graph of a wave-like curve with two peaks and a valley, showing multiple points with horizontal tangents.

Curve with multiple horizontal tangents.

not differentiable

note there is no place where \( f' = 0 \)

discontinuous

so no place where \( f' = 0 \)

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example

\( f(x) = e^{x^4} \) on \( [-2, 2] \)

verify Rolle's Theorem

\( f(x) \) is exponential function so is continuous and differentiable everywhere, so first two requirements are met.

\( f(-2) = f(2) \) ?

\( f(-2) = e^{(-2)^4} = e^{16} \)

\( f(2) = e^{(2)^4} = e^{16} \)

so, 3rd requirement is met.

Rolle's Theorem guarantees at least one place, \( -2 < x < 2 \), where the tangent line is horizontal.

Find it:

\( f'(x) = e^{x^4} \cdot 4x^3 = 4x^3 e^{x^4} \)

\( f' = 0 \rightarrow 4x^3 e^{x^4} = 0 \)

\( 4x^3 = 0, \quad e^{x^4} \neq 0 \) (exponential \( \neq 0 \))

\( x = 0 \)

\( f'(0) = 4(0) e^{(0)} = 0 \) → horiz. tangent line

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Rolle's Theorem doesn't help us find where \( f'(x) = 0 \), but it assures us there is at least such location so we can spend the time and effort to find it.

( if Rolle's Theorem says there isn't one, then there is no point to look for it ).

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An extension to Rolle's Theorem is called

The Mean Value Theorem

If \( f(x) \) is

Continuous on \( [a, b] \)
Differentiable on \( (a, b) \)

then there is at least one location \( c \), \( a < c < b \) where

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

secant line slope

there is at least one place where the tangent line slope is equal to the secant line slope through the two end points

Figure: Coordinate graph showing a curve between points (a, f(a)) and (b, f(b)) with a parallel tangent at c.

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Example

\( f(x) = \cos x \) on \( [0, \frac{\pi}{2}] \)

\( \cos x \) is continuous and differentiable everywhere so both requirements are met.

Slope of secant line through end points

\[ \frac{f(\frac{\pi}{2}) - f(0)}{\frac{\pi}{2} - 0} = \frac{\cos(\frac{\pi}{2}) - \cos(0)}{\frac{\pi}{2}} = \frac{0 - 1}{\frac{\pi}{2}} = -\frac{2}{\pi} \]

Mean Value Theorem says there is at least one \( c \) where \( f'(c) = -\frac{2}{\pi} \)

\[ f'(x) = -\sin x = -\frac{2}{\pi} \]\[ \sin x = \frac{2}{\pi} \]\[ x = \sin^{-1}(\frac{2}{\pi}) \approx 0.7 \]

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4.3 What the Derivative Tells Us (part 1)

Figure: A coordinate graph showing a smooth curve starting at (0,0), peaking at x=1, crossing x-axis at x=2, and dipping at x=3.

on interval \( 0 < x < 1 \) (or \( (0, 1) \)) the function is increasing

notice on that interval the tangent line slope is always positive

on interval \( 1 < x < 3 \) the function is decreasing

notice on that interval the tangent line slope is always negative

this tells us:

if \( f' > 0 \) on some interval, \( f \) is increasing
if \( f' < 0 \) on some interval, \( f \) is decreasing

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The First Derivative Test

There is a relative maximum at \( x=1 \) where \( f' = 0 \).

Notice \( f' > 0 \) before and \( f' < 0 \) after \( x=1 \).

There is a relative minimum at \( x=3 \) where \( f' = 0 \).

Notice \( f' < 0 \) before and \( f' > 0 \) after \( x=3 \).

This observation is the foundation of:

First Derivative Test

If \( x=c \) is a critical number (\( f'(c)=0 \) or \( f'(c) \) DNE) and if \( f'(x) \) changes from \( + \) to \( - \) across \( x=c \), then there is a relative max at \( x=c \).

If \( f'(x) \) changes from \( - \) to \( + \) across \( x=c \), then there is a relative min at \( x=c \).

If \( f'(x) \) does not change sign, then there is neither a relative max nor a relative min at \( x=c \).

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Example

\( f(x) = 3x^4 - 4x^3 \)

Find critical numbers: \( f' = 0 \) or \( f' \) DNE

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

\( f' = 0 \rightarrow 12x^3 - 12x^2 = 0 \)

\( 12x^2(x - 1) = 0 \rightarrow \) \( x=0, x=1 \) Critical numbers

\( f' \) DNE \( \rightarrow \) never, since \( f(x) \) is a polynomial, it is always differentiable.

Two critical numbers: \( x=0, x=1 \)

Draw a number line with these on it, then track sign of \( f' \).

Figure: A number line for f'(x) with critical points at 0 and 1. Signs are negative, negative, then positive.

Pick ANY \( x < 0 \) and check sign of \( f' \)

\( f'(-1) = 12(-1)^3 - 12(-1)^2 < 0 \)

Pick ANY \( x \) between 0 and 1 check sign of \( f' \)

\( f'(\frac{1}{2}) = 12(\frac{1}{2})^3 - 12(\frac{1}{2})^2 < 0 \)

Pick ANY \( x > 1 \) check sign of \( f' \)

\( f'(2) = 12(2)^3 - 12(2)^2 > 0 \)

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Analysis of Function Monotonicity and Extrema

Intervals of Increase and Decrease

Based on the derivative analysis, we can conclude the following about the behavior of the function \( f(x) \):

\( f(x) \) is decreasing on the intervals \( (-\infty, 0) \) and \( (0, 1) \).
\( f(x) \) is increasing on the interval \( (1, \infty) \).

First Derivative Test for Relative Extrema

\( f' \) changes from negative (\(-\)) to positive (\(+\)) across \( x = 1 \).

So there is a relative minimum at \( x = 1 \).

\( f' \) does not change sign across \( x = 0 \).

So there is neither a relative maximum nor a relative minimum at \( x = 0 \).