Find the derivative at each critical point and determine the local extreme valse

Question

Find the derivative at each critical point and determine the local extreme valses. Use deci appropriate values. Use decimal approximations when B) Critical Ptlderivativel Critical Pt derivative ExtremunVaue A) Critical P:lderivative Extremum/Value undefined min min lo Ilocal max 400,5 x=20 x=20 Cnitical Pt derivative Extremum Value x 20 local max D) C) Critical Pe Iderivative Extrermum Value x-0 x-25 max 005 ocal max undefined min local max 4005 1 15 x3-6x2+8,x1 A) B) Critical Pt. [derivative x--1 -defined|local min Extremum Value Crtical Pt. Iderivative Extremum/ Value Critical Pt 4 x=6.213 |local max 1-308 local max -3.08 D) Critical Pt Iderivative Extremum Value Critical Pt. alue local max local max 14 max- local min -3.08 local max -3.08

Explanation / Answer

13.

I = integral (x^n) = x^(n+1)/(n+1) + C

I = integral (x^-4 + (1/3)*x^(-1/2))

= x^(-4+1)/(-4+1) + (1/3)*x^(-1/2 + 1)/(-1/2 + 1) + C

= -x^-3/3 + (2/3)*x^(1/2) + C

= -1/3*x^3 + 2*x^(1/2)/3 + C

Correct option is C.

14.

I = integral (5t^2 + t/6)*dt

= 5*t^3/3 + t^2/(6*2) + C

= 5t^3/3 + t^2/12 + C

Correct option is C.

15.

I = integral [(x*x^(1/2) + x^(1/2))/x^2]*dt

I = integral [x^(3/2 - 2) + x^(1/2 - 2)]dt

I = integral [x^(-1/2) + x^(-3/2)]dt

= x^(-1/2 + 1)/(-1/2 + 1) + x^(-3/2 + 1)/(-3/2 + 1) + C

= 2*x^(1/2) - 2*x^(-1/2) + C

Correct option is C.

16.

d((8x+3)^4)/dx = 4*(8x + 3)^3*(8*1 + 0)

= 32*(8x + 3)^3

Yes

17.

y = (x^2*sin x)/2

dy/dx = (2x*sin x)/2 + (x^2*cos x)/2

dy/dx = x*sin x + (x^2/2)*cos x

No

Related Questions

Navigate

• Browse (All)
• Browse F
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.