Calculus problem help please?

Question

I'm supposed to graph this function considering the domain, critical points, regions where the function is increasing or decreasing, points of inflection, regions where the function is concave upward or concave downward, and intercepts where possible.

f(x)=x^5-15x^3

if anyone could please help me it would be greatly appreciated. Don't worry I don't expect anyone to type up a graph :) based on the other information I'd be able to draw it. Thank you

Answer 1

hi,is that an assignment? i also got one assignment like that, but with different function of course



domain:

- ~ < x < +~ [ since no restriction in the equation]

[ "~" is infinity sign]





x-intercept:

f(x) = 0

x^5 - 15x^3 =0

x= -3.87 || x=0 || x= 3.87

( -3.87,0) ( 0,0) ( 3.87,0)



y-intercept:

f(0) = (0)^5 - 15(0)^3 = 0

(0,0)





turning points:

f '(x)= 0

5x^4 - 45 x ^2=0

x= 3 || x= 0 || x=-3



f(-3) = 162 f(0)= 0 f(3) = -162

therefore the turning points are:

(-3, 162) & (0,0) & ( 3, -162)





Set up f'(x) test table:

interval | x<-3 | -33 |

f ' (x) | + | - | - | + |







function increasing:

f'(x) >0 [f'(x) is positive]

from test table, the interval of f(x) increasing:

(x<-3) U (x>3)

or to be specific

(- ~






function decreasing:

f'(x) <0 [f'(x) is negative]

from test table, the interval of f(x) increasing:

(-3




Points of inflection:

f " (x) = 0

20x^3 - 90x = 0

x = -2.12 || x=0 || x=2.12



f(-2.12) = 100.10 f(0)= 0 f(2.12) = -100.10

therefore the POI are:

( -2.12, 100.10) & (0,0) & (2.12, -100.10)





Set up f " (x) test table:

interval | x< -1.12 | -2.122.12 |

f " (x) | - | + | - | + |







concave up:

f " (x) >0 [f " (x) is positive]

from test table, the interval of f(x) increasing:

(-2.122.12)

or to be specific

(-2.12






concave down:

f " (x) <0 [f " (x) is negative]

from test table, the interval of f(x) increasing:

( x < -1.12 ) U ( 0
or to be specific

( -~ < x < -1.12 ) U ( 0








then, graph the function. good luck. (^-^)b

Answer 2

f (x) = x^3 - 12x + a million . . . the 1st spinoff set to 0 reveals turning or table sure factors f ' (x) = 3x^2 - 12 3x^2 - 12 = 0 3 * (x + 2) * (x - 2) = 0 x = 2 ... x = - 2 . . . the 2d spinoff evaluated at x = 2 and -2 determines if those factors are min, max, or neither. f ' ' (x) = 6x f ' ' (2) = 6*2 = 12 <== beneficial value shows x=2 is a interior of sight minimum f ' ' (-2) = 6*(-2) = -12 <== damaging value shows x=-2 is a interior of sight optimal a.) x = - 2 is a optimal, and x=2 is a minimum ... so x = - infinity to -2 is increasing x = -2 to +2 is lowering x = +2 to + infinity is increasing b.) f (-2) = (-2)^3 - 12*(-2) + a million = 17 f (2) = (2)^3 - 12*(2) + a million = - 15 c.) . . . the 2d spinoff set to 0 reveals inflection factors, or the place concavity adjustments 6x = 0 x = 0 <=== inflection factor x = - 2 is a optimal, so must be concave down concavity adjustments on the inflection factor(s) ... so x = - infinity to 0 is concave down x = 0 to + infinity is concave up

Answer 3

The critical points are the points where x and y are zero, and where the derivative is zero.



1) Set x = 0, and find y or f(x)

2) Set y or f(x) = 0, and solve for x.

3) Differentiate, set dy/dx = 0, and solve for x. Substitute these values into f(x) to find the corresponding y values.

4) Differentiate again to see where these points are maximum or minimum. Usually the function is decreasing after a maximum and increasing after a minimum.

Answer 4

looks tough lol