The graph of the continuous function g, the derivative of the function f, is shown above. the function g is piecewise linear for -5 ≤ x < 3 and g(x) = 2(x-4)^2 for 3 ≤ x ≤ 6.

(a) if f(1) = 3, what is the value of f(-5) ?

(b) evaluate \int_{0}^{6} g(x)\; dx .

(c) for -5 < x < 6, on what open intervals, if any, is the graph of f both increasing and concave up? give a reason for your answer.

(d) find the x-coordinate of each point of inflection of the graph of f. give a reason for your answer.

(this was on the no-calculator section of the recently-released ap calculus ab 2018 exam so i appreciate it if you tried to limit calculator usage)

(a)

f is the antiderivative of g



We calculate using geometry.

For x ∈ [-5,-1], g is a negative trapezoid. If the area of a trapezoid is 1/2(a+b)(h), then a = 4, b = 3, h = -3. The area here is therefore



For x ∈ [-1,0], g has no area between the curve and the x-axis. Therefore,

For x ∈ [0,1], g is a positive triangle. If the area of a triangle is 1/2bh, then b = 1 and h = +2. The area here is therefore



Thus



Hence



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(b)

For





Thus



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(c)

The graph of f is both increasing and concave up on the intervals 0 < x < 1 and 4 < x < 6 because the derivative of f, which is the function g graphed above, is positive and increasing.

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(d)

The graph of f has an inflection point at x = 4 because the derivative of f changes from decreasing to increasing at this point.

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