Understanding the Problem
To solve for g'(0), we first need to analyze the function g(x) defined as:
g(x) = ∫ from 1 to 3 f(x + t) dt
Here, the function f is continuous, and we know its values at specific points: f(1) = 5 and f(3) = 11.
Applying the Fundamental Theorem of Calculus
To find g'(x), we differentiate g(x) with respect to x:
g'(x) = d/dx [∫ from 1 to 3 f(x + t) dt]
Using Leibniz's rule, we get:
g'(x) = ∫ from 1 to 3 f'(x + t) dt
Now, we specifically want to find g'(0):
g'(0) = ∫ from 1 to 3 f'(0 + t) dt = ∫ from 1 to 3 f'(t) dt
Finding f'(t)
Since we only know the values of f at specific points, we can use the Mean Value Theorem for integrals. The mean value of f over the interval [1, 3] is given by:
Mean value = (f(1) + f(3)) / 2 = (5 + 11) / 2 = 8
This suggests that there's some average behavior of the function f on that interval.
Calculating g'(0)
For an approximation, we consider the average value of f over the interval. Since f is continuous, we assume it behaves linearly between these points for the sake of this calculation.
The derivative at the midpoint (2) can be approximated based on the values given:
f'(2) ≈ (f(3) - f(1)) / (3 - 1) = (11 - 5) / 2 = 3
Now, integrating over the interval:
g'(0) = ∫ from 1 to 3 f'(t) dt
This average leads us to estimate g'(0) ≈ 8, but considering the options, we refine our estimate to see that the closest value aligns with option (c) 6.
Conclusion
Thus, g'(0) is approximately:
Answer: (c) 6