In the following problem, suppose f(x) is continuous (and differentiable) function on the interval (0,1). Suppose also that for 0 < x<(1/4) f(x) is negative, for (1/4) <x<1 f(x) is positive, f(1/4)=0, f (2/3)=1, f ' (1/3)= 1, f ' (2/3) =3.
a. If the function G(x) is given by
G(x)= integral from 0 to x f(t) dt
what can you conclude about the maximum or minimum values of G on (0,1)?
b. What is the slope of the tangent line to G(x) at x=(2/3)?
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b. They tell you that the slope (f') there is 3.
To find the maximum or minimum values of G(x), we can consider the behavior of f(x) and the properties of integrals.
a. First, let's think about the properties of the integral. The integral of a function represents the accumulation of the function's values over a given interval. Since G(x) is defined as the integral of f(x) from 0 to x, it represents the total accumulated area under the curve of f(x) from 0 to x.
Given the information about f(x), we know that f(x) is negative for 0 < x < 1/4 and positive for 1/4 < x < 1. This means that the area under the curve of f(x) will be negative for 0 < x < 1/4 and positive for 1/4 < x < 1.
Since the integral accumulates these positive and negative areas, the maximum value of G(x) can be found when the area under the curve of f(x) is the largest, and the minimum value of G(x) can be found when the area under the curve of f(x) is the smallest.
To find the maximum or minimum values of G(x), we need to consider the behavior of the function f(x) between 0 and 1. The given information tells us that f(1/4) = 0, which means that the area under the curve of f(x) is zero at x = 1/4. This is a point where G(x) reaches an extremum, either a maximum or a minimum value.
To determine whether it is a maximum or minimum, we can look at the behavior of f(x) around x = 1/4. We know that f(x) is negative for 0 < x < 1/4 and positive for 1/4 < x < 1. This indicates that the area under the curve of f(x) is increasing as we move from 0 to 1/4 and then decreasing as we move from 1/4 to 1.
Since the area under the curve of f(x) is increasing before x = 1/4 and decreasing after x = 1/4, we can conclude that G(x) has a minimum value at x = 1/4.
b. To find the slope of the tangent line to G(x) at x = 2/3, we need to differentiate G(x) with respect to x.
The derivative of G(x) with respect to x is given by the Fundamental Theorem of Calculus:
G'(x) = f(x)
This means that G'(x) is equal to the original function f(x).
From the given information, we know that f'(1/3) = 1 and f'(2/3) = 3. Since G'(x) = f(x), this implies that G'(1/3) = 1 and G'(2/3) = 3.
Therefore, the slope of the tangent line to G(x) at x = 2/3 is 3.
To summarize:
a. The minimum value of G(x) occurs at x = 1/4.
b. The slope of the tangent line to G(x) at x = 2/3 is 3.