Graphs for Calculus:: Derivatives and Integrals from Graphs Flashcards (38 Cards) — Study Online

Curriculum Overview

Learn derivatives and integrals by reading graphs. 38 flashcards, each with a real graph on the front: a function and its derivative, tangent lines, concavity and inflection points, where a function is not differentiable (corners, cusps, vertical tangents, jumps), definite integrals from areas, accumulation functions and the Fundamental Theorem, the Mean Value Theorem, Rolle, absolute extrema and linear approximation. Every answer is checked and explained. For AP Calculus AB/BC and college Calculus 1.

Topics & Key Concepts

derivative integral tangent line slope increasing and decreasing local maximum local minimum concavity inflection point differentiability cusp area under a curve Fundamental Theorem of Calculus accumulation function Mean Value Theorem Rolle linear approximation AP Calculus Calculus 1

Sample Flashcard Questions & Answers

Showing 8 of 38 cards

Question #1 Active Recall

The graph of \(f\) (blue) and the graph of its derivative \(f'\) (red) are shown. Where is \(f\) increasing and decreasing, and where does \(f\) have a local maximum or a local minimum? Explain using \(f'\).

Answer & Explanation:

\(f'\) is positive for \(x < -1\) and for \(x>1\), so \(f\) is **increasing** on \((-\infty,-1)\) and \((1,\infty)\). \(f'\) is negative on \((-1,1)\), so \(f\) is **decreasing** there.

\(f'=0\) at \(x=-1\) and \(x=1\). At \(x=-1\), \(f'\) changes from \(+\) to \(-\): a **local maximum** (\(f(-1)=2\)). At \(x=1\), \(f'\) changes from \(-\) to \(+\): a **local minimum** (\(f(1)=-2\)). Here \(f(x)=x^3-3x\) and \(f'(x)=3x^2-3\).

Question #2 Active Recall

The graph of \(f(x)=x^2\) (blue) and the graph of its derivative \(f'\) (red) are shown. Write the equation of \(f'\), and say what the red graph tells you about the slope of \(f\) at \(x=-1\), \(x=0\) and \(x=2\).

Answer & Explanation:

\(f'(x)=2x\), a line through the origin with slope 2.

At \(x=-1\): \(f'(-1)=-2\), so \(f\) is falling with slope \(-2\). At \(x=0\): \(f'(0)=0\), a horizontal tangent (the vertex). At \(x=2\): \(f'(2)=4\), so \(f\) is rising with slope 4.

Question #3 Active Recall

The graph of \(f(x)=\sin x\) (blue) and the graph of its derivative \(f'\) (red) are shown. Write \(f'(x)\). Where in \([0, 2\pi]\) is \(f'(x)=0\), and what happens to \(f\) there?

Answer & Explanation:

\(f'(x)=\cos x\).

\(f'=0\) at \(x=\dfrac{\pi}{2}\) and \(x=\dfrac{3\pi}{2}\) (about 1.57 and 4.71). At \(x=\pi/2\), \(f'\) changes from \(+\) to \(-\): \(f\) has a **local maximum** (\(f=1\)). At \(x=3\pi/2\), \(f'\) changes from \(-\) to \(+\): \(f\) has a **local minimum** (\(f=-1\)).

Question #4 Active Recall

The graph of \(f(x)=\ln x\) (blue) and the graph of its derivative \(f'\) (red) are shown. Write \(f'(x)\) and explain why \(f\) is always increasing.

Answer & Explanation:

\(f'(x)=\dfrac{1}{x}\).

For every \(x>0\) (the whole domain of \(\ln x\)), \(\dfrac1x>0\), so the red graph stays above the x-axis and \(f\) is increasing everywhere. The slope is large near \(x=0\) (the blue graph is steep there) and gets smaller and smaller as \(x\) grows (it flattens), but never reaches 0. At \(x=1\) the slope is \(f'(1)=1\).

Question #5 Active Recall

The graph of \(f'\), the **derivative** of \(f\), is shown. Where is \(f\) increasing? Where does \(f\) have a local maximum and a local minimum?

Answer & Explanation:

The graph shown is \(f'(x)=(x+1)(x-2)\), which is positive for \(x < -1\) and for \(x>2\) and negative on \((-1,2)\).

So \(f\) is **increasing** on \((-\infty,-1)\) and \((2,\infty)\) and **decreasing** on \((-1,2)\). At \(x=-1\), \(f'\) changes from \(+\) to \(-\): a **local maximum**. At \(x=2\), \(f'\) changes from \(-\) to \(+\): a **local minimum**.

Question #6 Active Recall

The graph of \(f'\), the **derivative** of \(f\), is shown. \(f'(0)=0\). Does \(f\) have a local extremum at \(x=0\)? What happens at \(x=3\)?

Answer & Explanation:

**No extremum at \(x=0\).** The graph of \(f'\) touches the x-axis at 0 but stays below it on both sides, so \(f'\) does not change sign: \(f\) is decreasing before and after \(x=0\) (a flat spot).

**At \(x=3\)**, \(f'\) changes from negative to positive, so \(f\) has a **local minimum**. The graph shown is \(f'(x)=x^2(x-3)\); the double zero at 0 is why the sign does not change.

Question #7 Active Recall

The graph of \(f'\), the **derivative** of \(f\), is shown (a straight line). Where is \(f\) increasing? Where does \(f\) have a local maximum? Is \(f\) concave up or concave down? Why?

Answer & Explanation:

\(f'(x)=2-x\) is positive for \(x < 2\) and negative for \(x>2\), so \(f\) is **increasing** on \((-\infty,2)\) and **decreasing** on \((2,\infty)\), with a **local maximum at \(x=2\)**.

\(f'\) is a decreasing line (slope \(-1\)), so \(f''=-1 < 0\) everywhere: \(f\) is **concave down** everywhere. (For example \(f(x)=2x-\tfrac{x^2}{2}\), a downward parabola.)

Question #8 Active Recall

The graph of \(f(x)=x^2\) (blue) and its tangent line at \(x=1\) (red, with a dot at the point of tangency) are shown. Find the slope of the tangent line and its equation.

Answer & Explanation:

Slope: \(f'(1)=2\) (since \(f'(x)=2x\)). The point of tangency is \((1,1)\).

Equation: \(y-1=2(x-1)\), that is \(y=2x-1\).

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