Trig Review:General Shape of Functions
Worksheet Solutions
#1) A critical point is any value that causes f ‘(x) = 0 or f ‘(x) to not exist
f ‘(x) = 0 when sin x = 0f ‘(x) = DNE when cos x = 0
Over the interval of 0 to 4πcos x = 0 at x = π/2 + nπ… π/2, 3π/2, 5π/2, 7π/2
Critical Points at nπ/2
#2) Increasing when f ‘(x) > 0
sin (x) = 0 at x = nπsin (3x) = 0 at x = nπ/3
sin starts at zero and goes up to 1–sin and therefore –6sin will start at zero and go down
– + – + – + – + –
#3) Decreasing when f ‘(x) < 0
sin (x) = 0 at x = nπsin (x/3) = 0 at x = nπ/(1/3) = 3nπ
sin starts at zero and goes up to 1–sin and also –(1/3)sin will start at zero and go down
– +
#4) Max is when f ‘ goes from + to –
cos (x) = 0 at x = π/2 + nπcos (x + π/3) = 0 at x = π/2 + nπ – π/3cos ( + π/3) = 0 at x = π/6 + nπ
x = 0 2cos(0 + π/3) = 2cos(π/3) = 2 * .5 = 1 = positive
+ – + – +
#5) Min is when f ‘ goes from – to +
cos (x) = 0 at x = π/2 + nπcos (3x) = 0 at x = (π/2 + nπ)/3cos (3x) = 0 at x = π/6 + 2nπ/6
x = 0 8cos(4*0) = 8 *1 = 8 = positive
+ – + – + – + – +
#6) POI is when f “ = 0 and changes signs
sin (x) = 0 at x = nπsin (πx) = 0 at x = nπ/πsin (πx) = 0 at x = n
x = 0.5 -2π2 sin(π * .5) = -2π2 * 1 = negative
– + – +
Note that the range in the problem does not include 0 and 4 (uses < & >)
#7) Concave Up when f “ > 0
sin (x) = 0 at x = nπsin (x/2) = 0 at x = nπ/(1/2)sin (πx) = 0 at x = 2nπ
x = π (-1/2)sin(π/2) = (-1/2) * 1 = -1/2 = Negative
– +
#8) Concave Up when f “ < 0
sin (x) = 0 at x = nπsin (x – π/4) = 0 at x = nπ + π/4
x = 0 -5sin(0 - π/4) = -5 * -1/√2 = Positive
+ – + – +
#35) Period AmplitudeA 4 2B 2 1C π 2D 2 2E π 1
#32) A, only numbers inside the trig function affect the period of a trig function
#37) f (x) = ex sinx = 0 when either ex = 0 or sin x = 0ex never equals zerosin x = 0 at x = nπ. This interval includes: 0, π, 2πTotal of 3 zeros, D