This question has been regraded.

If
𝑓
is a differentiable function so that
𝑓
(
3
)
=
2
and
𝑓
′
(
3
)
=
1
3
, find the equation of the line tangent to
𝑦
=
𝑓
(
𝑥
)
at the point
(
3
,
𝑓
(
3
)
)
.

Question

This question has been regraded.

If 
𝑓
 is a differentiable function so that 
𝑓
(
3
)
=
2
 and 
𝑓
′
(
3
)
=
1
3
, find the equation of the line tangent to 
𝑦
=
𝑓
(
𝑥
)
 at the point 
(
3
,
𝑓
(
3
)
)
.

BlackTom AI · Accepted Answer

We start by identifying the key data given for the tangent line problem. We are told that f is differentiable and that f(3) = 2 and f'(3) = 1/3. The tangent line to y = f(x) at the point (3, f(3)) uses the point (3, 2) and has slope equal to f'(3), which is 1/3.

Using the point-slope form of the equation of a line: y - y1 = m(x - x1), with (x1, y1) = (3, 2) and m = 1/3, we get:
y - 2 = (1/3)(x - 3).

Expanding and simplifying: y = (1/3)x - 1 + 2 = (1/3)x + 1.

Thus, the tangent line to y = f(x) at x = 3 is y = (1/3)x + 1.

Note on the provided answer options: no explicit list of answer choices was given in the prompt, so there are no alternative statements to evaluate here. The derivation above shows why the line y = (1/3)x + 1 satisfies both the slope condition and passes through the given point.

This question has been regraded. If 𝑓 is a differentiable function so that 𝑓 ( 3 ) = 2 and 𝑓 ′ ( 3 ) = 1 3 , find the equation of the line tangent to 𝑦 = 𝑓 ( 𝑥 ) at the point ( 3 , 𝑓 ( 3 ) ) .单项选择题

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