The first derivative of the function f is defined by f'(x) = (x2 +1)sin(3x - 1) for -1.5 < x < 1.5. On which of the following intervals is the graph of f concave up?Responses(−1.5, −1.341) and (−0.240, 0.964)(−1.5, −1.341) and (−0.240, 0.964)(−1.341, −0.240) and (0.964, 1.5)(−1.341, −0.240) and (0.964, 1.5)(−0.714, 0.333) and (1.381, 1.5)(−0.714, 0.333) and (1.381, 1.5)(−1.5, −0.714) and (0.333, 1.381)

For what interval of x-values is the curve y = f (x) concave up?

The table gives values for a polynomial function 𝑔 at selected values of 𝑥. If 𝑎<𝑏, then 𝑔𝑎>𝑔𝑏 for all 𝑎 and 𝑏 in the interval 3<𝑥<7. Which of the following could be true about the graph of 𝑔 on the interval 3<𝑥<7 ?ResponsesThe graph of 𝑔 is concave down because the function is decreasing, and the average rate of change over equal-length input-value intervals is increasing.The graph of  g  is concave down because the function is decreasing, and the average rate of change over equal-length input-value intervals is increasing.The graph of 𝑔 is concave up because the function is decreasing, and the average rate of change over equal-length input-value intervals is increasing.The graph of  g  is concave up because the function is decreasing, and the average rate of change over equal-length input-value intervals is increasing.The graph of 𝑔 is concave down because the function is decreasing, and the average rate of change over equal-length input-value intervals is decreasing.The graph of  g  is concave down because the function is decreasing, and the average rate of change over equal-length input-value intervals is decreasing.The graph of 𝑔 is concave up because the function is decreasing, and the average rate of change over equal-length input-value intervals is decreasing.

Use the given graph to estimate the value of each derivative. (Round all answers to one decimal place.)(a) f ' (-3) (b) f ' (-2) (c) f ' (-1) (d) f ' (0) (e) f ' (1)

The graph of a function f is given in the figure.A curve is shown on the x y coordinate plane. It begins at the point (−2, −1), goes up and to the right, passes through the approximate point (−1, −0.2), and passes through the negative x-axis at the approximate point (−0.8, 0). It then continues up and right, passes through the positive y-axis at the point (0, 1), and reaches a high point at (1, 3). It then goes down and right, passes through the points (2, 2) and (3, 1), and ends at the approximate point (4, 0.5).(a)Find the value of f(1).(b)Estimate the value of f(−1).(c)For what values of x is f(x) = 1? (Enter your answers as a comma-separated list.) (d)Estimate the value of x such that f(x) = 0.x = (e)State the domain and range of f. (Enter your answers in interval notation.)domain range (f)On what interval is f increasing? (Enter your answer using interval notation.)

determine\:the\:intervals\:of\:concavity\:and\:the\:points\:of\:inflection\:for\:the\:curve\:y=3x^5-40x^3+3x-20