Sketch curves using derivative information

Curve sketching from derivative information involves relating the sign of $f'(x)$ to intervals of increase and decrease, critical points, and local maxima or minima, and relating the sign of $f''(x)$ to concavity and possible points of inflection. Combined with intercepts, domain restrictions, end behavior, and asymptotes where appropriate, these features support a coherent graph of a standard polynomial, rational, exponential, logarithmic, or trigonometric function; parametric and implicit curves and more advanced higher-derivative analysis are not included.

Detailed Explanation: Sketch curves using derivative information

The key connections are:

If $f'(x)>0$ , the graph is increasing.
If $f'(x)<0$ , the graph is decreasing.
A change in $f'(x)$ from positive to negative gives a local maximum.
A change in $f'(x)$ from negative to positive gives a local minimum.
If $f''(x)<0$ , the graph is concave down.
If $f''(x)>0$ , the graph is concave up.

Use these facts together with intercepts and end behavior to sketch the curve.

Example

Sketch the curve

f(x)=x^3-3x^2+2.

1. Find the intercepts

For the $y$ -intercept, set $x=0$ :

f(0)=2.

So the $y$ -intercept is $(0,2)$ .

For the $x$ -intercepts, solve $f(x)=0$ :

x^3-3x^2+2=0.

Factoring gives

x^3-3x^2+2=(x-1)(x^2-2x-2).

Thus,

x=1,\qquad x=1-\sqrt{3},\qquad x=1+\sqrt{3}.

The graph crosses the $x$ -axis at these three points.

2. Use $f'(x)$ to find increasing and decreasing intervals

Differentiate:

f'(x)=3x^2-6x=3x(x-2).

The critical numbers occur where $f'(x)=0$ :

3x(x-2)=0 \quad\Longrightarrow\quad x=0,\ 2.

Test the sign of $f'(x)$ on the intervals determined by $0$ and $2$ :

\begin{array}{c|c|c} \text{Interval} & \text{Sign of }f'(x) & \text{Behavior}\\ \hline (-\infty,0) & + & \text{increasing}\\ (0,2) & - & \text{decreasing}\\ (2,\infty) & + & \text{increasing} \end{array}

Therefore, the graph increases, then decreases, then increases again.

Find the function values at the critical numbers:

f(0)=2,\qquad f(2)=8-12+2=-2.

Since $f'(x)$ changes from positive to negative at $x=0$ , there is a local maximum at

(0,2).

Since $f'(x)$ changes from negative to positive at $x=2$ , there is a local minimum at

(2,-2).

3. Use $f''(x)$ to determine concavity

Differentiate again:

f''(x)=6x-6.

Set $f''(x)=0$ :

6x-6=0 \quad\Longrightarrow\quad x=1.

Check the sign of $f''(x)$ :

If $x<1$ , then $f''(x)<0$ , so the graph is concave down.
If $x>1$ , then $f''(x)>0$ , so the graph is concave up.

The concavity changes at $x=1$ , so there is an inflection point. Its $y$ -coordinate is

f(1)=1-3+2=0.

Thus, the point of inflection is

(1,0).

4. Determine the end behavior

The leading term is $x^3$ . Therefore,

x\to-\infty \implies f(x)\to-\infty,

and

x\to\infty \implies f(x)\to\infty.

So the graph goes down on the far left and up on the far right.

5. Put the information together

Important points and features are:

$x$ -intercepts: $(1-\sqrt3,0)$ , $(1,0)$ , and $(1+\sqrt3,0)$
$y$ -intercept and local maximum: $(0,2)$
Local minimum: $(2,-2)$
Inflection point: $(1,0)$
Increasing on $(-\infty,0)$ and $(2,\infty)$
Decreasing on $(0,2)$
Concave down for $x<1$
Concave up for $x>1$

Sketch a smooth cubic curve that starts low on the left, rises to the local maximum $(0,2)$ , falls through the inflection point $(1,0)$ to the local minimum $(2,-2)$ , and then rises to the right.

Learn by doing: Sketch curves using derivative information

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Critical Points - Derivative Chart to Graph

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