Rich boi

Use the graph of f(t) = 2t + 2 on the interval [–1, 4] to write the function F(x), where f of x equals the integral from 1 to x of f of t dt. F(x) = x2 + 3x F(x) = x2 + 2x – 12 F(x) = x2 + 2x – 3 F(x) = x2 + 4x – 8

Lim(x>inf)=2000x^4/200x^4= 10

f(x) and g(x) are a differentiable function for all reals and h(x) = g[f(5x)]. The table below gives selected values for f(x), g(x), f '(x), and g '(x). Find the value of h'(1). x 1 2 3 4 5 6 f(x) 0 3 2 1 2 0 g(x) 1 3 2 6 5 0 f '(x) 3 2 1 4 0 2 g '(x) 1 5...

The function is continuous on the interval [10, 20] with some of its values given in the table above. Estimate the average value of the function with a Right Hand Sum Approximation, using the intervals between those given points. x 10 12 15 19 20 f(x) –2...

Find the area under the graph of f(x) = e-ln(x) on the interval [1, 2].

(05.02)If the integral from 2 to 5 of f of x, dx equals 4 and the integral from 2 to 7 of f of x, dx equals negative 3, then what is the value of integral from 5 to 7 of f of x, dx?

Thank you

f(x) has a relative maximum at x = 0. f(x) is concave upwards for all x. f(x) has a point of inflection at x = 0. f(x) passes through the origin.

Show that f(x) = x3 and g(x) = 200x3 grow at the same rate.

I only need the 2nd one

–12 –7 7 –1

o ok