Calculus Reference

Curve Analysis

The first derivative describes slope and motion; the second derivative describes how that slope changes.

Solved Examples

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Derivative Tests

QuestionTestConclusion
Where can extrema occur?Find $f'(x)=0$ or where $f'$ is undefined.These are critical numbers in the domain of $f$.
Where is $f$ increasing?Check the sign of $f'$.$f'>0$ means increasing; $f'<0$ means decreasing.
What is the curve doing?Check the sign of $f''$.$f''>0$ means concave up; $f''<0$ means concave down.

Find Critical Points

Quadratic Expression

Find the critical point of $f(x)=x^2-4x+3$.

Solution Steps
  1. Find the first derivative.
  2. Set the derivative equal to zero and solve for $x$.
  3. Evaluate the original function at the critical number.
  4. State the critical point: $(2,-1)$.
$$f'(x)=2x-4$$$$2x-4=0\quad\Rightarrow\quad x=2$$$$f(2)=4-8+3=-1$$$$\text{critical point: }(2,-1)$$

Find Increasing and Decreasing Intervals

Sign of the First Derivative

Classify $f(x)=x^2-4x+3$ on either side of $x=2$.

Solution Steps
  1. Test the derivative at a point left of $x=2$.
  2. Test the derivative at a point right of $x=2$ and state the intervals.
$$f'(1)=-2<0\quad\Rightarrow\quad\text{decreasing on }(-\infty,2)$$$$f'(3)=2>0\quad\Rightarrow\quad\text{increasing on }(2,\infty)$$

Find Concavity and Inflection

Cubic Expression

Find the concavity of $g(x)=x^3$.

Solution Steps
  1. Find the first and second derivatives.
  2. Use the sign of the second derivative for $x<0$.
  3. Use the sign of the second derivative for $x>0$.
  4. State the inflection point: $(0,0)$.
$$g'(x)=3x^2,\qquad g''(x)=6x$$$$x<0:\ g''(x)<0\Rightarrow\text{concave down}$$$$x>0:\ g''(x)>0\Rightarrow\text{concave up}$$$$\text{inflection point: }(0,0)$$