Answer to Question #297066 in Calculus for Ram

Question #297066

<e> find the turning points and point of inflection on the curve, y=x5-5x4+5x3-1




1
Expert's answer
2022-02-25T08:50:55-0500
y=5x420x3+15x2y'=5x^4-20x^3+15x^2

Find the ctitical number(s)


y=0=>5x420x3+15x2=0y'=0=>5x^4-20x^3+15x^2=0

5x2(x24x+3)=05x^2(x^2-4x+3)=0

5x2(x1)(x3)=05x^2 (x-1)(x-3)=0

x1=0,x2=1,x3=3x_1=0, x_2=1, x_3=3

Find the second derivative


y=20x360x2+30xy''=20x^3-60x^2+30x

y(0)=0y''(0)=0

The function ff has no turning point at x=0.x=0.



y(1)=2060+30=10<0y''(1)=20-60+30=-10<0

The function ff has turning point at x=1.x=1. The function ff has a local maximum at x=1.x=1.



y(3)=540540+90=90>0y''(3)=540-540+90=90>0

The function ff has turning point at x=3.x=3. The function ff has a local minimum at x=3.x=3.


Find the inflection point(s)


y=0=>20x360x2+30x=0y''=0=>20x^3-60x^2+30x=0

10x(2x26x+3)=010x(2x^2-6x+3)=0

5x(2x33)(2x3+3)5x(2x-3-\sqrt{3})(2x-3+\sqrt{3})

Point of inflection at x=0,x=332,x=3+32.x=0, x=\dfrac{3-\sqrt{3}}{2}, x=\dfrac{3+\sqrt{3}}{2}.


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment