Can someone help me with these math problems?

Question

Determine and classify the extrema of
f(x)= 2x5 - 5x4

detrmine wether the critical point is a max. min . or inflection.

y= 3x3 - 9x - 5 x= -1
y= 2x3 - x5 x=0

p.s. the # to the right of the x are exponents

Answer 1

Hi,

f(x)= 2x^5 - 5x^4

Relative maximum (0,0)
Relative minimum (2,-16)
Inflection Point (1.5,-10.125)
As x -> - ∞ , y -> - ∞
As x -> ∞ , y -> ∞

y= 3x³ - 9x - 5 at x= -1
The point (-1,1) is a relative maximum.
First derivative is 9x² - 9 which solves to x = 1 and x = -1 as min and max respectively. This can be easily verified on a graphing calculator.

y= 2x³ - x^5 at x=0
(0,0) is an inflection point between concave down and concave up sections. The second derivative gives inflection points at x = 0, ±√(15)/5

I hope that helps!! :-)

Answer 2

1) f(x) = x^4(2x-5)

so f(x) = 0 when x=0, 5/2
as x tends to infinity f(x) tends to negative infinity
as x tends to negative infinity f(x) tends to infinity


f'(x) = 10x^4 - 20x^3
= 10x^3(x-2)

so stationary points at x =0 and x = 2

f''(x) = 40x^3 - 60x^2
= 20x^2 (2x - 3)

x = 0 => f''(x) = 0 therefore x = 0 gives a point of inflexion

x = 2 => f''(x) > 0 therefore a minimum point

2) y = 3x^3 - 9x - 5

y' = 9x^2 - 9 = 9 (x^2 -1)

y' = 0 when x^2 - 1 = 0 ie x^2 = 1 => x = +/-1

y'' = 18x

for x = -1 y'' <0 therefore is a maximum point

3) y = 2x^3 - x^5
= x^3 (2 - x^2)

y' = 6x^2 - 5x^4
= x^2 (6 - 5x^2)

y' = 0 when x^2 = 0 or 6 - 5x^2 = 0
so x^2 = 0 , 6/5
so x = 0 , +/- sqrt 6/5 gives the stationary points

y'' = 12x - 20x^3
= 4x(3 - 5x^2)

for x = 0 => y'' = 0 the stationary point is a point of inflexion

Answer 3

f(x) = 2x^5 - 5x^4

Extrema occur when df/dx = 0 so

df/dx = 0 = 10x^4 - 20x^3
0 = 10x - 20 = x - 2 ----> x=2

Classify by the value of the second derivative.

d^f/dx^2 = 40x^3 - 60x^2 evaluate at x = 2

d^2f/dx^2 = 320 - 240 > 0 so x = 2 is a minimum

Note sure what the other equatoins two are since they are both equal to constants, but you fllow the same procedure.

Answer 4

for the next 2 equation do a d^2y.dx^2 if it is then sub the value of x to detemine +Positive or negative result if Postive it is minima if negative then maima if zero then nothing can be derive