Derivatives

TopicDerivatives

Title

The slope of a cubic from its secants

ExampleW05.1

Step1 / 5

Mode

Given

For $f(x) = x^3 - 3x^2 + 2x + 1$ , find the slope of the secant from $x = 2$ over the runs $h = 1$ , $0.1$ and $0.01$ , then the slope they close on, $f'(2)$ , and the tangent line at $x = 2$ .

$f(x) = x^3 - 3x^2 + 2x + 1$
the point $a = 2$
the runs $h = 1$ , $0.1$ , $0.01$

Each run: the secant's slope

\frac{f(2 + h) - f(2)}{h}

$h$	$f(2 + h)$	$\frac{f(2 + h) - f(2)}{h}$
1
0.1
0.01
$\to 0$

Read the problem

A secant from

x = 2

over a run

h

rises

f(2 + h) - f(2)

, so its slope is

\frac{f(2 + h) - f(2)}{h}

. As

h

shrinks, the secant turns onto the tangent, and its slope onto

f'(2)

Sim

Secants closing on the tangent

Slide $P$ along the curve to move the tangent, or $Q$ to change the run $h$ . The table's secants, over $h$ , a tenth of it and a hundredth, close on the slope the red pen rings.

Assumed: supplied here, not in the Materials: $a$ tuned from $0$ to $2.5$ in steps of $0.05$ · $h$ tuned from $0.1$ to $1$ in steps of $0.05$

y

x

Fig.

f(x)

, its tangent at

P

and the secants towards

Q

. Slide

P

Q

along the curve.

a

2.00

h

1.00

$h$	$f(a + h)$	$\frac{f(a + h) - f(a)}{h}$
1	7.0000	6.0000
0.1	1.2310	2.3100
0.01	1.0203	2.0301
$h \to 0$	1.0000	2.0000

The secants close on $f'(a)$ = 2.0000: the tangent at $P$ is y = 2.0000x − 3.0000

Worked examples

The slope of a cubic from its secants

Read the problem

Secants closing on the tangent