This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: 3 x ( x4) =3 x = x4 = x = x = 4 Now to find the points, you must find the corresponding yvalues for these xvalues by substituting back into x=0 x=4 Thus, the tangent line to the graph of f is horizontal at the points (0, 0) and (4, 32). f ( x ) = x 3 + 6 x 2 . f (0) = 3 + 6 ⋅ 2 f (0) = f (4) = 4 3 + 6 ⋅ 4 2 f (4) = 32 Graph for this example Write the equation of the tangent line at a given point on the graph Write the equation of the tangent line to the graph of at the point (3,9) First find the slope using m = . For this function . At the point (3,9), f x ( 29 = x 2 , ′ f x ( 29 = 2 x ( 29 ( 29 3 2 3 6 m f ′ === ( 29 f x ′ Now use m = 6 and the point (3,9) to write the equation for the tangent line ( 29 ( 29 y x y x y x y x =   = + = = 9 6 3 9 6 3 9 6 18 6 9 Thus the tangent line to the graph of at the point (3,9) is the line f x ( 29 = x 2 , y x = 6 9...
View
Full Document
 Spring '09
 Calculus, Derivative, Vector Space, Continuous function

Click to edit the document details