Single-Variable-Calculus · Unit 7 · Video 4 · Interactive Practice

Quadratic Approximation: Adding the Second-Derivative Term

IKey Formulas

Formula	Name	What you need
$f(x) \approx f(x_0) + f'(x_0)(x - x_0)$	Linear approximation	Value and slope at $x_0$
$f(x) \approx f(x_0) + f'(x_0)(x - x_0) + \dfrac{f''(x_0)}{2}(x - x_0)^2$	Quadratic approximation	One more number: $f''(x_0)$
$f(x) \approx f(0) + f'(0)\,x + \dfrac{f''(0)}{2}\,x^2 \quad (x \approx 0)$	Base point $0$	$f(0)$ , $f'(0)$ , $f''(0)$
$\sin x \approx x \qquad \cos x \approx 1 - \tfrac{1}{2}x^2 \qquad e^x \approx 1 + x + \tfrac{1}{2}x^2$ $\ln(1 + x) \approx x - \tfrac{1}{2}x^2 \qquad (1 + x)^r \approx 1 + rx + \tfrac{r(r-1)}{2}x^2$	Quadratic catalog	Valid for $x \approx 0$

Key Insight: The linear part stays exactly as it was; the quadratic term only adds $\tfrac{f''(x_0)}{2}(x - x_0)^2$ . It turns the tangent line into the best-fit parabola and removes most of the error near $x_0$ — but no term makes the formula reliable far from $x_0$ .

IIVisualization 1 — From Tangent Line to Best-Fit Parabola

Adding $\tfrac{f''(0)}{2}x^2$ keeps the tangent line's value and slope and bends it to follow the curve near $0$ .

f(x) x 0.50

IIIVisualization 2 — How Much Closer Is ln(1 + x)?

For $\ln(1 + x)$ , how much closer does $x - \tfrac{1}{2}x^2$ land than the linear answer $x$ ?

x 0.10

IVVisualization 3 — Which Parabola Hugs cos x?

Among the parabolas $y = 1 + c\,x^2$ through $(0, 1)$ , which one fits $\cos x$ best near $0$ ?

c 0.00

💡 The winning coefficient is $c = \tfrac{f''(0)}{2}$ with $f''(0) = -\cos 0 = -1$ ; why the factor is exactly $\tfrac{1}{2}$ is the subject of the next lecture.

VQuiz Questions

Problem 1 · Estimate $e^{0.2}$

Given: $e^x \approx 1 + x + \tfrac{1}{2}x^2$ for $x \approx 0$ — estimate $e^{0.2}$ .

✅ Correct!

1 + 0.2 + \tfrac{1}{2}(0.04) = 1.22

, against the true

e^{0.2} \approx 1.2214

❌ That is the linear approximation.

1 + x

gives

1.2

; the quadratic term adds

\tfrac{1}{2}x^2 = 0.02

❌ The factor $\tfrac{1}{2}$ went missing. The quadratic term is

\tfrac{f''(0)}{2}x^2 = \tfrac{1}{2}(0.2)^2 = 0.02

, not

(0.2)^2 = 0.04

❌ The linear term went missing. Keep

1 + x

and add the quadratic term on the end.

❌ Not quite. Substitute

x = 0.2

into

1 + x + \tfrac{1}{2}x^2

Show solution

With $x = 0.2$ :

e^{0.2} \approx 1 + 0.2 + \tfrac{1}{2}(0.2)^2 = 1 + 0.2 + \tfrac{1}{2}(0.04) = 1 + 0.2 + 0.02 = 1.22.

The true value is $e^{0.2} \approx 1.22140$ , so the error drops from about $0.021$ (linear, $1.2$ ) to about $0.0014$ .

Problem 2 · The Missing Term

Given: $f(x) = \sin x$ — find its quadratic approximation near $x = 0$ .

✅ Correct!

f''(0) = -\sin 0 = 0

, so there is no

x^2

term: the quadratic approximation of

\sin x

is the linear one.

❌ The second derivative is $-\sin x$ , and that is $0$ at $x = 0$ . A coefficient of

-1

would come from

-\cos 0

-\tfrac{1}{(1+0)^2}

— those belong to

\cos x

and

\ln(1 + x)

❌ Evaluate $f''$ at the base point.

f''(x) = -\sin x

and

\sin 0 = 0

, so the quadratic coefficient is

0

, not

\pm\tfrac{1}{2}

❌ That is the approximation of $\cos x$ . For

\sin x

the value at

0

0

and the slope is

\cos 0 = 1

, so the linear part is

x

❌ Not quite. Compute

f(0)

f'(0)

and

f''(0)

for

\sin x

Show solution

Derivatives: $f(x) = \sin x$ , $f'(x) = \cos x$ , $f''(x) = -\sin x$ . At $0$ :

f(0) = 0, \qquad f'(0) = \cos 0 = 1, \qquad f''(0) = -\sin 0 = 0.

\sin x \approx 0 + 1 \cdot x + \frac{0}{2}x^2 = x.

The quadratic term vanishes, which is why $\sin x \approx x$ is such a good approximation: it is also the quadratic approximation.

Problem 3 · A Base Point Other Than Zero

Given: $f(x) = \sqrt{x}$ with base point $x_0 = 4$ — use the quadratic approximation to estimate $\sqrt{4.2}$ .

What is $f''(4)$ ?

What is the quadratic estimate of $\sqrt{4.2}$ ?

✅ Correct!

2 + \tfrac{1}{4}(0.2) - \tfrac{1}{64}(0.2)^2 = 2.049375

, and

\sqrt{4.2} \approx 2.049390

— off by about

0.000015

❌ Check the sign.

\sqrt{x}

bends downward:

f''(x) = -\tfrac{1}{4}x^{-3/2}

is negative for every

x \gt 0

❌ Check the coefficient. Differentiating

\tfrac{1}{2}x^{-1/2}

gives

\tfrac{1}{2}\cdot\big(-\tfrac{1}{2}\big)x^{-3/2} = -\tfrac{1}{4}x^{-3/2}

, and

4^{-3/2} = \tfrac{1}{8}

❌ That is $f'(4)$ .

f'(x) = \tfrac{1}{2\sqrt{x}}

gives

\tfrac{1}{4}

; differentiate once more for

f''

❌ That is the linear estimate.

2 + \tfrac{1}{4}(0.2) = 2.05

; the quadratic term still has to be added.

❌ The factor $\tfrac{1}{2}$ went missing. The term is

\tfrac{f''(4)}{2}(0.2)^2 = -\tfrac{1}{64}(0.04) = -0.000625

, not

-\tfrac{1}{32}(0.04) = -0.00125

❌ The quadratic term has the sign of $f''(4)$ .

f''(4) \lt 0

, so the correction is subtracted from

2.05

❌ Differentiate twice.

f'(x) = \tfrac{1}{2}x^{-1/2}

; apply the power rule again, then put

x = 4

❌ Assemble the three pieces.

f(4) + f'(4)(0.2) + \tfrac{f''(4)}{2}(0.2)^2

Show solution

Step 1 — Derivatives. $f(x) = x^{1/2}$ , $f'(x) = \tfrac{1}{2}x^{-1/2}$ , $f''(x) = -\tfrac{1}{4}x^{-3/2}$ . At $x_0 = 4$ :

f(4) = 2, \qquad f'(4) = \frac{1}{2 \cdot 2} = \frac{1}{4}, \qquad f''(4) = -\frac{1}{4} \cdot \frac{1}{8} = -\frac{1}{32}.

Step 2 — Assemble with $x - x_0 = 0.2$ :

\sqrt{4.2} \approx 2 + \frac{1}{4}(0.2) + \frac{-1/32}{2}(0.2)^2 = 2 + 0.05 - \frac{0.04}{64} = 2.05 - 0.000625 = 2.049375.

The true value is $\sqrt{4.2} \approx 2.0493902$ : the linear estimate $2.05$ is off by about $0.0006$ , the quadratic by about $0.000015$ .

Problem 4 · A Negative Power

Given: $(1 + x)^r \approx 1 + rx + \tfrac{r(r-1)}{2}x^2$ — find the quadratic approximation of $\dfrac{1}{1 + x}$ near $0$ , then estimate $\dfrac{1}{1.1}$ .

Which is the quadratic approximation of $\dfrac{1}{1 + x}$ ?

What does it give for $\dfrac{1}{1.1}$ ?

✅ Correct! With

r = -1

the coefficient is

\tfrac{(-1)(-2)}{2} = 1

, so

\tfrac{1}{1.1} \approx 1 - 0.1 + 0.01 = 0.91

, against the true

0.90909\ldots

❌ Check the sign of $r(r - 1)$ . With

r = -1

r(r - 1) = (-1)(-2) = +2

: two negatives make the

x^2

coefficient positive.

❌ Divide by $2$ .

r(r-1) = 2

, but the coefficient is

\tfrac{r(r-1)}{2} = 1

❌ $\dfrac{1}{1 + x} = (1 + x)^{-1}$ , so $r = -1$ . The linear term is

rx = -x

❌ That is the linear estimate.

1 - x

gives

0.9

; the quadratic term adds

x^2 = 0.01

❌ That uses $2x^2$ . The coefficient of

x^2

\tfrac{r(r-1)}{2} = 1

, so the term is

0.01

, not

0.02

❌ Write the function as a power.

\tfrac{1}{1 + x} = (1 + x)^{-1}

; put

r = -1

into the catalog formula.

❌ Substitute $x = 0.1$ into your approximation

1 - x + x^2

Show solution

Step 1 — Identify $r$ . $\dfrac{1}{1 + x} = (1 + x)^{-1}$ , so $r = -1$ .

Step 2 — Coefficients. $rx = -x$ and

\frac{r(r-1)}{2} = \frac{(-1)(-2)}{2} = 1, \qquad\text{so}\qquad \frac{1}{1 + x} \approx 1 - x + x^2.

Step 3 — Evaluate. $1.1 = 1 + 0.1$ , so $x = 0.1$ :

\frac{1}{1.1} \approx 1 - 0.1 + 0.01 = 0.91.

The true value is $0.90909\ldots$ : the error falls from about $0.009$ (linear, $0.9$ ) to about $0.0009$ .

Solved: 0 / 4