Ap calculus ab 2010 scoring guidelines (form
b. question 2

Alright

question 2 is as follows (see screenshot for problem and answers)

a.
horizontal tangent is where the slope is 0 or where the derivitive of g(x) is 0
so solve for g'(x)=0
we are given g'(x)=sin(x+1/x)
set to zero
0=sin(x+1/x)
we solve and graph and find whre it is in the bounds given and find taht
x=0.163 and x=0.359 (always round to 3 decimal places unless told otherwise)

b. concave down is where 2nd derivitive is negative
first find 0's
just graph it and see where the graph is below 0
that is in the interval (0.1295,0.2227), hmm, seeems they want 4 decimal places,

c. tangent, so we need a point and a sllope to use the point slope form
well, we can integegrate
given that g(1)=2
remember

so, we want to find g(0.3)
we do


add 2 to both sides and evaluate the integral
we get that g(0.3)=1.546
so point slope form
the slope is g'(3)=-0.472161
so point slope form would be
y-1.546=-0.472(x-0.3)
if we wanted it in terms of y, easy
y=1.546-0.472(x-0.3)

d.
easy way is we can see if it is concave down or up
that's 2nd derivitive
g''(x)>0 in the range from 0.3 to 1, so therefor the tangent line is below the graph