---
title: "Visualizing $f$, $f'$, $f''$"
subtitle: "Math 119 — Applied Calculus for Data Analysis (BYU-Idaho)"
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---

# How to use this worksheet

**Draw first, then run the code.** The goal is to predict the shape of $f'$ and $f''$ from the shape of $f$ — running the cells before sketching defeats the point.

For each example below:

1. **Sketch $f(x)$** on paper using the formula given.
2. Mark the **zeros of $f$** with **filled (black) dots** on the $x$-axis.
3. On each interval, decide whether the slope of $f$ is **positive (+)** or **negative (−)** and write the sign above it. Then mark each **inflection point** of $f$ with an **open dot** on the $x$-axis.
4. **Hand-sketch $f'$** on the same axes in a different color (e.g. red). Use your $+/-$ signs to predict where $f'$ is above/below the axis, and your open dots to predict where $f'$ has a peak or valley.
5. Repeat the slope/inflection analysis on **$f'$**, then hand-sketch **$f''$** (a third color, e.g. blue).

After each step, **run the matching code cell** to check your work. Colors stay consistent across examples: $f$ in black, $f'$ in red, $f''$ in blue.

## Setup — run once

```{r setup}
plot_derivs <- function(x, f, fp = NULL, fpp = NULL,
                        zeros = NULL, inflections = NULL,
                        ylim = NULL, main = "") {
  ys <- f(x)
  if (!is.null(fp))  ys <- c(ys, fp(x))
  if (!is.null(fpp)) ys <- c(ys, fpp(x))
  if (is.null(ylim)) ylim <- range(ys, finite = TRUE)
  plot(x, f(x), type = "l", lwd = 2, col = "black",
       ylim = ylim, ylab = "y", main = main)
  abline(h = 0, col = "gray80")
  if (!is.null(fp))  lines(x, fp(x),  col = "red",  lwd = 2)
  if (!is.null(fpp)) lines(x, fpp(x), col = "blue", lwd = 2)
  if (!is.null(zeros))       points(zeros, rep(0, length(zeros)), pch = 16, cex = 1.3)
  if (!is.null(inflections)) points(inflections, rep(0, length(inflections)), pch = 1, cex = 1.6, lwd = 2)
  legend_lab <- "f"; legend_col <- "black"
  if (!is.null(fp))  { legend_lab <- c(legend_lab, "f'");  legend_col <- c(legend_col, "red") }
  if (!is.null(fpp)) { legend_lab <- c(legend_lab, "f''"); legend_col <- c(legend_col, "blue") }
  legend("topright", legend = legend_lab, col = legend_col, lty = 1, lwd = 2, bg = "white")
}
```

# Example A — $f(x) = x^2$

Start simple. Sketch the parabola first, then check each step.

**Step 1 — zeros of $f$ (filled black dots).** $f(x) = x^2 = 0$ only at $x = 0$.

```{r ex-a-step1}
f <- function(x){ x^2 }
x <- seq(-3, 3, 0.01)
plot_derivs(x, f,
            zeros = c(0),
            ylim  = c(-2, 9),
            main  = "f(x) = x^2")
```

**Step 2 — inflection points + overlay $f'$.** $f'(x) = 2x$. $f''(x) = 2$ — never zero, so $f$ has **no inflection points** (no open dots).

```{r ex-a-step2}
fp <- function(x){ 2*x }
plot_derivs(x, f, fp = fp,
            zeros       = c(0),
            inflections = NULL,
            ylim        = c(-7, 9),
            main        = "f and f'")
```

**Step 3 — overlay $f''$.** $f''(x) = 2$ is a flat horizontal line, always positive → $f$ is concave up everywhere.

```{r ex-a-step3}
fpp <- function(x){ rep(2, length(x)) }
plot_derivs(x, f, fp = fp, fpp = fpp,
            zeros       = c(0),
            inflections = NULL,
            ylim        = c(-7, 9),
            main        = "f, f', f''")
```

# Example B — $f(x) = x^3$

Same single zero at the origin, but $f$, $f'$, $f''$ interact differently.

**Step 1 — zero at $x = 0$ (triple zero — the curve flattens against the axis).**

```{r ex-b-step1}
f <- function(x){ x^3 }
x <- seq(-2, 2, 0.01)
plot_derivs(x, f,
            zeros = c(0),
            ylim  = c(-8, 8),
            main  = "f(x) = x^3")
```

**Step 2 — inflection at $x = 0$ (open dot) + overlay $f'$.** $f'(x) = 3x^2 \ge 0$ always. $f''(x) = 6x$ changes sign at $x = 0$, so $x = 0$ IS an inflection point.

**Important:** $f'(0) = 0$ but it's NOT a max or min — it's a *horizontal inflection*. A zero of $f'$ only gives a max or min when $f'$ also changes sign there.

```{r ex-b-step2}
fp <- function(x){ 3*x^2 }
plot_derivs(x, f, fp = fp,
            zeros       = c(0),
            inflections = c(0),
            ylim        = c(-8, 12),
            main        = "f and f'")
```

**Step 3 — overlay $f''$.** $f''(x) = 6x$.

```{r ex-b-step3}
fpp <- function(x){ 6*x }
plot_derivs(x, f, fp = fp, fpp = fpp,
            zeros       = c(0),
            inflections = c(0),
            ylim        = c(-8, 12),
            main        = "f, f', f''")
```

# Example C — $f(x) = x^4$

Looks similar to $x^2$ — but hides a subtle trap about $f''$.

**Step 1.**

```{r ex-c-step1}
f <- function(x){ x^4 }
x <- seq(-2, 2, 0.01)
plot_derivs(x, f,
            zeros = c(0),
            ylim  = c(-2, 16),
            main  = "f(x) = x^4")
```

**Step 2.** $f'(x) = 4x^3$. $f''(x) = 12x^2 \ge 0$ always — $f''(0) = 0$ but no sign change → **NOT** an inflection point.

```{r ex-c-step2}
fp <- function(x){ 4*x^3 }
plot_derivs(x, f, fp = fp,
            zeros       = c(0),
            inflections = NULL,
            ylim        = c(-20, 20),
            main        = "f and f'")
```

**Step 3 — overlay $f''$.**

```{r ex-c-step3}
fpp <- function(x){ 12*x^2 }
plot_derivs(x, f, fp = fp, fpp = fpp,
            zeros       = c(0),
            inflections = NULL,
            ylim        = c(-20, 20),
            main        = "f, f', f''")
```

# Example D — $f(x) = e^x$

The function that is its own derivative. Predict the plot before running.

**Step 1 — no zeros** ($e^x > 0$ for every real $x$).

```{r ex-d-step1}
f <- function(x){ exp(x) }
x <- seq(-2, 3, 0.01)
plot_derivs(x, f,
            zeros = NULL,
            ylim  = c(0, 20),
            main  = "f(x) = e^x")
```

**Step 2 — $f'(x) = e^x = f$.** The red curve overlays exactly on the black curve.

```{r ex-d-step2}
fp <- function(x){ exp(x) }
plot_derivs(x, f, fp = fp,
            zeros       = NULL,
            inflections = NULL,
            ylim        = c(0, 20),
            main        = "f and f' (identical)")
```

**Step 3 — $f''(x) = e^x$ again.** All three curves coincide.

```{r ex-d-step3}
fpp <- function(x){ exp(x) }
plot_derivs(x, f, fp = fp, fpp = fpp,
            zeros       = NULL,
            inflections = NULL,
            ylim        = c(0, 20),
            main        = "f, f', f'' (all three coincide)")
```

# Example E — $f(x) = \ln(x)$

Restricted domain ($x > 0$), monotonic increasing, concave down.

**Step 1 — zero of $f$ at $x = 1$.**

```{r ex-e-step1}
f <- function(x){ log(x) }       # log() in R is the natural log
x <- seq(0.1, 5, 0.01)            # start past 0 to avoid -Inf
plot_derivs(x, f,
            zeros = c(1),
            ylim  = c(-3, 2),
            main  = "f(x) = ln(x)")
```

**Step 2.** $f'(x) = 1/x$ (always positive on the domain). $f''(x) = -1/x^2$ (always negative). No inflection points.

```{r ex-e-step2}
fp <- function(x){ 1/x }
plot_derivs(x, f, fp = fp,
            zeros       = c(1),
            inflections = NULL,
            ylim        = c(-3, 5),
            main        = "f and f'")
```

**Step 3 — $f''$ negative everywhere on the domain → $f$ concave down everywhere.**

```{r ex-e-step3}
fpp <- function(x){ -1/x^2 }
plot_derivs(x, f, fp = fp, fpp = fpp,
            zeros       = c(1),
            inflections = NULL,
            ylim        = c(-3, 5),
            main        = "f, f', f''")
```

# Example F — $f(x) = x^3 - 3x$

Now the zeros of $f$, $f'$, and $f''$ don't line up at all. This is closer to what you'll see on the [Graphing Using Signs worksheet](https://chaz-clark.github.io/m119-site/assets/docs/GraphingUsingSigns.pdf) — a function with **one local max**, **one local min**, and **one inflection point**, all at different $x$ values.

**Step 1 — zeros of $f$.** $f(x) = x(x^2 - 3) = 0$ at $x = 0, \pm\sqrt{3} \approx \pm 1.732$.

```{r ex-f-step1}
f <- function(x){ x^3 - 3*x }
x <- seq(-2.5, 2.5, 0.01)
plot_derivs(x, f,
            zeros = c(-sqrt(3), 0, sqrt(3)),
            ylim  = c(-4, 4),
            main  = "f(x) = x^3 - 3x")
```

**Step 2 — inflection at $x = 0$ + overlay $f'$.** $f'(x) = 3x^2 - 3 = 3(x^2 - 1)$, so $f'$ has zeros at $x = \pm 1$ — that's a **local max at $x = -1$** and a **local min at $x = +1$**. $f''(x) = 6x$ has a sign change at $x = 0$ (inflection).

```{r ex-f-step2}
fp <- function(x){ 3*x^2 - 3 }
plot_derivs(x, f, fp = fp,
            zeros       = c(-sqrt(3), 0, sqrt(3)),
            inflections = c(0),
            ylim        = c(-6, 8),
            main        = "f and f'")
```

Notice how the **red** $f'$ crosses zero exactly at $\pm 1$ — the peak and valley of $f$. The open dot at $x = 0$ sits at the bottom of $f'$'s own valley (the minimum of $f'$).

**Step 3 — overlay $f''$.**

```{r ex-f-step3}
fpp <- function(x){ 6*x }
plot_derivs(x, f, fp = fp, fpp = fpp,
            zeros       = c(-sqrt(3), 0, sqrt(3)),
            inflections = c(0),
            ylim        = c(-15, 15),
            main        = "f, f', f''")
```

The **blue** $f''$ crosses zero at the open dot ($x = 0$), confirming it's the inflection point.

# Example G — $f(x) = x^4 - 4x^2$

A "W-shaped" quartic with **two local mins**, **one local max**, and **two inflection points**. This is the polynomial with the richest critical-point structure in the worksheet.

**Step 1 — zeros of $f$.** $f(x) = x^2(x^2 - 4) = 0$ at $x = 0$ (double zero) and $\pm 2$.

```{r ex-g-step1}
f <- function(x){ x^4 - 4*x^2 }
x <- seq(-2.5, 2.5, 0.01)
plot_derivs(x, f,
            zeros = c(-2, 0, 2),
            ylim  = c(-5, 8),
            main  = "f(x) = x^4 - 4x^2")
```

**Step 2 — two inflection points + overlay $f'$.** $f'(x) = 4x^3 - 8x = 4x(x^2 - 2)$, with zeros at $x = 0, \pm\sqrt{2}$:

- $x = -\sqrt{2}$ is a **local min** of $f$
- $x = 0$ is a **local max** of $f$
- $x = +\sqrt{2}$ is another **local min** of $f$

$f''(x) = 12x^2 - 8$ has zeros at $x = \pm\sqrt{2/3} \approx \pm 0.816$ — **two** inflection points.

```{r ex-g-step2}
fp <- function(x){ 4*x^3 - 8*x }
plot_derivs(x, f, fp = fp,
            zeros       = c(-2, 0, 2),
            inflections = c(-sqrt(2/3), sqrt(2/3)),
            ylim        = c(-10, 10),
            main        = "f and f'")
```

There are **two** open dots now (the two inflection points of $f$). The red $f'$ crosses zero three times — at the two valleys and the central peak of $f$.

**Step 3 — overlay $f''$.**

```{r ex-g-step3}
fpp <- function(x){ 12*x^2 - 8 }
plot_derivs(x, f, fp = fp, fpp = fpp,
            zeros       = c(-2, 0, 2),
            inflections = c(-sqrt(2/3), sqrt(2/3)),
            ylim        = c(-12, 12),
            main        = "f, f', f''")
```

The **blue** $f''$ is the parabola $12x^2 - 8$. It crosses zero at both open-dot $x$-values — confirming both inflection points.

# Going further

Once the pattern feels natural:

- Compare Examples F and G to the [Graphing Using Signs worksheet](https://chaz-clark.github.io/m119-site/assets/docs/GraphingUsingSigns.pdf). Pick a graph from that PDF and apply the same workflow — sketch first, then verify with `plot_derivs()` if you can write down a formula.
- Sketch $f'''$ as well — for $x^n$ each derivative drops a power; for $e^x$ everything stays the same; for $\ln(x)$ the signs alternate.
- Try a **quintic** like $f(x) = x^5 - 5x^3 + 4x$ — multiple zeros, multiple extrema, multiple inflections. Predict the shape of $f'$ before plotting.
- Combine your examples: $f(x) = x^2 \ln(x)$ or $f(x) = e^x - x$ — much richer behavior.

The goal: **read the shape of $f'$ from the shape of $f$**, and the shape of $f''$ from the shape of $f'$, without computing anything algebraically.