Interpretation — instantaneous rate of change, slope of tangent: Question bank

True or false — justify

The derivative

f^{'} (a)

is a ratio (a fraction) at the point

x = a

False. It is the limit of a ratio

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

; at

h = 0

the ratio itself is

\frac{0}{0}

, undefined. The derivative is the single number that ratio approaches, not any actual fraction.

If the average rate of change over

[a, b]

5

, then

f^{'} (a) = 5

False in general. The average is one secant slope over a whole interval; the instantaneous rate is the limit

h \to 0

. They coincide only for straight lines, where the slope never changes.

A tangent line can only touch the curve at exactly one point.

False. "Tangent" means matching slope locally via

f^{'} (a)

, not "touches once." The tangent to

y = x^{3}

at the origin,

y = 0

, passes straight through the curve there.

f

is continuous at

a

, then

f^{'} (a)

must exist.

False. Continuity is necessary but not sufficient.

f (x) = ∣ x ∣

is continuous at

0

yet has a corner, so no single tangent slope exists. See Differentiability and continuity.

f^{'} (a)

exists, then

f

is continuous at

a

True. Differentiability forces continuity: a jump or hole would make

f (a + h) - f (a)

not shrink to

0

, so the limit defining

f^{'} (a)

couldn't exist.

f^{'} (a) = 0

means the function is not changing anywhere near

a

False. It means the instantaneous rate at that exact point is zero (a flat tangent). The function can be changing rapidly just to either side — a peak or valley has

f^{'} = 0

but is far from constant.

The secant slope always overestimates the tangent slope.

False. Whether the secant is steeper or shallower than the tangent depends on the curve's bending (concavity) and the sign of

h

. There's no universal direction. See Average rate of change.

Both limit forms

lim_{h \to 0} \frac{f ( a + h ) - f ( a )}{h}

and

lim_{x \to a} \frac{f ( x ) - f ( a )}{x - a}

give the same value when the derivative exists.

True. They are the same expression with the substitution

x = a + h

(so

h = x - a

). As

h \to 0

x \to a

; nothing else changes.

A negative derivative means the function's graph is below the

x

-axis.

False. The sign of $f^{'}$ describes slope direction (decreasing), not the sign of

f

. A curve can sit high above the axis while sloping downward, giving a positive value but a negative derivative.

Spot the error

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

h = 0

equals

\frac{f ( a ) - f ( a )}{0} = \frac{0}{0} = 0

The error:

\frac{0}{0}

is indeterminate, not

0

. You must simplify algebraically (cancel the

h

) before taking the limit — "Simplify Before Limit."

"Velocity is distance over time, so instantaneous velocity is

\frac{s ( 2 )}{2}

That's average velocity from

t = 0

t = 2

, not instantaneous. Instantaneous velocity is

lim_{h \to 0} \frac{s ( 2 + h ) - s ( 2 )}{h} = s^{'} (2)

. See Velocity and acceleration.

"The tangent line at

x = a

y = f^{'} (a) x

Missing the point. A line needs a point and a slope:

y - f (a) = f^{'} (a) (x - a)

. Their version forgets the curve passes through

(a, f (a))

, not the origin.

∣ x ∣

has derivative

0

x = 0

because the graph reaches its lowest point there."

Being a minimum doesn't guarantee a derivative. The one-sided slopes are

- 1

(left) and

+ 1

(right); they disagree, so the two-sided limit — and hence

f^{'} (0)

— does not exist.

"As

h \to 0

the secant becomes the tangent, so the secant slope equals the tangent slope for small

h

For any actual small

h \neq = 0

the secant slope is only approximately the tangent slope. Equality holds only in the limit, not at any real nonzero

h

"Since

f^{'} (a)

uses two points

a

and

a + h

, it's a property of an interval."

The two points are just scaffolding; the limit collapses them to the single point

a

. The derivative is a local property at one point, not an interval property.

Why questions

Why do we build a secant first instead of computing the tangent slope directly?

A secant uses two genuine points, so its rise-over-run is honestly computable with no division by zero. The tangent needs a limiting process to reach, which we get by shrinking the secant.

Why must we cancel the

h

before taking the limit?

The

\frac{0}{0}

trouble lives entirely in that shared factor of

h

. Cancelling removes the illegal division while

h \neq = 0

, leaving an expression whose limit we can evaluate by substitution.

Why can the same number

f^{'} (a)

mean both "slope of tangent" and "rate of change"?

The tangent is the best straight-line model of the curve near

a

; its slope is

\frac{rise}{run} = \frac{change in output}{change in input}

, which is exactly a rate of change. Same idea, two costumes.

Why does the derivative require a two-sided limit?

A well-defined tangent slope must be approached identically from both directions. If left and right give different limits (a corner), there's no single line grazing the curve, so no derivative.

Why is average rate not enough to describe a curved graph's behaviour?

On a curve the slope changes point to point, so one average over

[a, b]

hides all that variation. Only the limit

h \to 0

pins down the rate at each specific instant. See Derivative as a function.

Why does letting

a

vary turn

f^{'} (a)

into a whole new function

f^{'} (x)

The limit recipe works at every point where it exists, outputting a slope for each input. Collecting all those outputs gives a function — the derivative as a function, which shortcuts like the Power Rule then compute directly.

Edge cases

At a sharp corner (like

∣ x ∣

0

), what exactly fails?

The one-sided difference quotients approach different values (

- 1

and

+ 1

), so the two-sided limit doesn't exist and there is no unique tangent slope.

At a vertical tangent (e.g.

y = x^{1/3}

x = 0

), does

f^{'} (0)

exist?

No finite value: the difference quotient blows up to

+ \infty

. The tangent is vertical (undefined slope), so

f^{'} (0)

does not exist as a real number even though the curve is smooth-looking.

f

is a constant function, what is

f^{'} (a)

everywhere, and why?

Zero. Every difference quotient is

\frac{c - c}{h} = 0

for all

h \neq = 0

, so the limit is

0

— a horizontal, flat tangent, matching "not changing."

f

is a straight line

y = m x + b

, how do average and instantaneous rates compare?

They are identical and equal to

m

everywhere. The difference quotient simplifies to

m

for every

h

, so no limiting subtlety arises — this is the one case where average equals instantaneous.

What happens to the difference quotient at a jump discontinuity?

h \to 0

across the jump,

f (a + h) - f (a)

does not shrink to

0

, so dividing by a tiny

h

sends the quotient to

\pm \infty

. The derivative does not exist — discontinuity kills differentiability.

Connections

Average rate of change — the secant slope every "true/false" here contrasts against.
Limits — formal definition — why "approaches" is rigorous, not hand-waving.
Derivative as a function — the "letting $a$ vary" edge/why item.
Power Rule — the shortcut these limits justify.
Differentiability and continuity — the corner, jump, and vertical-tangent cases.
Velocity and acceleration — the average-vs-instantaneous velocity trap.