y ′ = 5 x 4 − 20 x 3 + 15 x 2 y'=5x^4-20x^3+15x^2 y ′ = 5 x 4 − 20 x 3 + 15 x 2 Find the ctitical number(s)
y ′ = 0 = > 5 x 4 − 20 x 3 + 15 x 2 = 0 y'=0=>5x^4-20x^3+15x^2=0 y ′ = 0 => 5 x 4 − 20 x 3 + 15 x 2 = 0
5 x 2 ( x 2 − 4 x + 3 ) = 0 5x^2(x^2-4x+3)=0 5 x 2 ( x 2 − 4 x + 3 ) = 0
5 x 2 ( x − 1 ) ( x − 3 ) = 0 5x^2 (x-1)(x-3)=0 5 x 2 ( x − 1 ) ( x − 3 ) = 0
x 1 = 0 , x 2 = 1 , x 3 = 3 x_1=0, x_2=1, x_3=3 x 1 = 0 , x 2 = 1 , x 3 = 3 Find the second derivative
y ′ ′ = 20 x 3 − 60 x 2 + 30 x y''=20x^3-60x^2+30x y ′′ = 20 x 3 − 60 x 2 + 30 x
y ′ ′ ( 0 ) = 0 y''(0)=0 y ′′ ( 0 ) = 0 The function f f f has no turning point at x = 0. x=0. x = 0.
y ′ ′ ( 1 ) = 20 − 60 + 30 = − 10 < 0 y''(1)=20-60+30=-10<0 y ′′ ( 1 ) = 20 − 60 + 30 = − 10 < 0 The function f f f has turning point at x = 1. x=1. x = 1. The function f f f has a local maximum at x = 1. x=1. x = 1.
y ′ ′ ( 3 ) = 540 − 540 + 90 = 90 > 0 y''(3)=540-540+90=90>0 y ′′ ( 3 ) = 540 − 540 + 90 = 90 > 0 The function f f f has turning point at x = 3. x=3. x = 3. The function f f f has a local minimum at x = 3. x=3. x = 3.
Find the inflection point(s)
y ′ ′ = 0 = > 20 x 3 − 60 x 2 + 30 x = 0 y''=0=>20x^3-60x^2+30x=0 y ′′ = 0 => 20 x 3 − 60 x 2 + 30 x = 0
10 x ( 2 x 2 − 6 x + 3 ) = 0 10x(2x^2-6x+3)=0 10 x ( 2 x 2 − 6 x + 3 ) = 0
5 x ( 2 x − 3 − 3 ) ( 2 x − 3 + 3 ) 5x(2x-3-\sqrt{3})(2x-3+\sqrt{3}) 5 x ( 2 x − 3 − 3 ) ( 2 x − 3 + 3 ) Point of inflection at x = 0 , x = 3 − 3 2 , x = 3 + 3 2 . x=0, x=\dfrac{3-\sqrt{3}}{2}, x=\dfrac{3+\sqrt{3}}{2}. x = 0 , x = 2 3 − 3 , x = 2 3 + 3 .
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