If
𝑓
(
16
)
=
14
,
𝑓
′
is continuous, and
∫
0
16
𝑓
′
(
𝑡
)
𝑑
𝑡
=
17
, what is the value of
𝑓
(
0
)
?

Question

If 
𝑓
(
16
)
=
14
, 
𝑓
′
 is continuous, and 
∫
0
16
𝑓
′
(
𝑡
)
𝑑
𝑡
=
17
, what is the value of 
𝑓
(
0
)
?

BlackTom AI · Accepted Answer

The Fundamental Theorem of Calculus connects a function’s derivative to the net change of the function: for any differentiable f with continuous derivative on [a,b], ∫_a^b f'(t) dt = f(b) − f(a). Continuity of f′ guarantees the integral exists and that the FTC applies.

1. Relevant knowledge points and concepts
- Fundamental Theorem of Calculus (Part 2): ∫_a^b f'(t) dt = f(b) − f(a).
- Continuity of f′ ensures f is differentiable and the integral is valid.
- Algebraic manipulation to solve for the unknown initial value f(0).

2. Complete reasoning step by step
- Apply the FTC with a = 0 and b = 16:
  ∫_0^16 f′(t) dt = f(16) − f(0).
- The problem gives ∫_0^16 f′(t) dt = 17 and f(16) = 14. Substitute:
  17 = 14 − f(0).

3. Calculations with intermediate steps
- 17 = 14 − f(0)
- Rearranging: −f(0) = 17 − 14 = 3
- Multiply by −1: f(0) = −3

4. Why the correct answer is correct
- The result directly follows from the FTC: the definite integral equals the net change f(16) − f(0); solving that equation yields f(0) = −3. All given conditions (value at 16 and the integral of f′) are sufficient.

5. Why the other options are incorrect
- 31: would correspond to f(0) = 14 − (−17), which misinterprets the sign of the integral; inconsistent with FTC.
- “There is not enough information”: incorrect because the FTC gives exactly the relation needed to find f(0).
- −2: does not satisfy 14 − (−2) = 16, not equal to the given integral 17.

6. Conclusion
- Therefore f(0) = −3, which matches the provided correct answer.

If 𝑓 ( 16 ) = 14 , 𝑓 ′ is continuous, and ∫ 0 16 𝑓 ′ ( 𝑡 ) 𝑑 𝑡 = 17 , what is the value of 𝑓 ( 0 ) ?Single choice

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