5 Steps to a 5 AP Calculus BC 2019

176 STEP 4. Review the Knowledge You Need to Score High

[−.5π, π] by [−4, 4]

Figure 9.1-4

y=−3 sin 2x;

dy

dx

=−3[cos(2x)]2=−6 cos(2x)

Slope of tangent at

(

x=

π

2

)

:

dy

dx

∣∣

∣

∣x=π/ 2 =−6 cos [2(π/2)]=−6 cosπ=^6.

Point of tangency atx=

π

2

,y=−3 sin(2x)

=−3 sin[2(π/2)]=−3 sin(π)= 0.

Therefore,

(

π

2

,0

)

is the point of tangency.

Equation of tangent:y− 0 =6(x−π/2) ory= 6 x− 3 π.

Example 2

If the liney= 6 x+ais tangent to the graph ofy= 2 x^3 , find the value(s) ofa.

Solution:

y= 2 x^3 ;

dy

dx

= 6 x^2 .(See Figure 9.1-5.)

[−2, 2] by [−6, 6]

Figure 9.1-5

The slope of the liney= 6 x+ais 6.

Sincey= 6 x+ais tangent to the graph ofy= 2 x^3 , thus

dy

dx

=6 for some values ofx.