Please help and remember that the graph is f and what your trying to find is the g(x)

Question

Graph: https://us-static.z-dn.net/files/d5d/53c5936d080f0d9eaf90caac5fae33ad.jpg

Question 1: https://us-static.z-dn.net/files/de6/e92ee03d331706b90f794be7ed3d9b72.jpg

Question 2: https://us-static.z-dn.net/files/d56/8b35873b43ef56be3a29f8c22aed2061.jpg

Answer 1

I agree with Samuel Z's responses, except for the following:

1. a. g(6)=(1/2)·2·4 + (1/2)(2)(-4)=0, g(8)=(1/2)·2·4+(1/2)(4)(-4)=-4.

b. g(x) increases, means g'(x)=f(x)>0, therefore in (-4,4) interval; likewise, g(x) decreases in interval (4,8). You would not include the endpoints of each interval because at those points f(x)=0.

c. g(x) is concave up, which means g"(x)>0, that is f'(x)>0 over the (-4,2) and (6,8) open intervals; g(x) decreases in (4,8); therefore, their open intersection (6,8) is the open interval that g(x) is concave up and decreasing;

d. no change

e. The slope of the tangent line at g(6) is g'(6)=f(6)=-4. The value of g at g(6)=0. Therefore, the equation of the tangent line to g at x=6 is y=-4(x-6)+0, or y=-4x+24;

f. no change

g. I'm happy to send you an image as well. :)

2. h(x)=g(3x)

a. no change

b. Tangent line of h(x) is g'(3x) = f(3x). At x=1, f(3x)=f(3⋅1)=f(3)=2. And g(3⋅1)=1. So the equation of the tangent line to h at x=1 is y=2(x-3)+1, or y=2x-5.

c. no change

I hope this helps!!!

Answer 2

1. a. Integral of aa function is the AREA value of the shape made up by function graph, x-axis and interval ending lines: therefore, g(-4)=-(1/4)·π·(4)²-2·4 =-8-4π, g(0)=-2·4=-8, g(4)=(1/2)·2·4=4, g(6)=2·(1/2)·2·4=-8, g(8)=(1/2)·2·4+(1/2)·4·4=-12.

b. g(x) increases, means g'(x)=f(x)>0, therefore in [-4,4] interval; likewise, g(x) decreases in interval [4,8].

c. g(x) is concave up, which means g"(x)<0, that is f'(x)<0 in (2,6) open interval; g(x) decreases in [4,8]; therefore, their open intersection (4,6) is the open interval that g(x) is concave up and decrease;

d. The inflection points of g(x) mean where g"(x)=0, that is f'(x)=0, at x=2 and 6.

e. Tangent line of g(x) at x=6, that is f(x) at x=6, which are y=-2x+8 and y=2x-16;

f. Based on the conclusion of a section, range of g(x) is [-8-4π, 4].

g. Please message me and I will send you the image.

1. a. lim_x→2[g(3x)/(x-2)] = lim_x→2[g'(3x)] = lim_x→2[f(3x)] = -4.

b. h(x)=g(3x), tangent line of h(x) is g'(3x) = f(3x), at x=1, it is f(x) at x=3, that is y=-2x+8.

c. h'(x) = g'(3x) = f(3x), so h'(0) = f(0)=4.