Solve f(x)=x^3-6x^2+15: Help Needed

  • Thread starter hectorubie
  • Start date
In summary, the given function is f(x)=x^3-6x^2+15, and its derivative is 3x^2-12x. The critical points are x=0 and x=4, and the function is positive from (-∞,0) and (4,∞), and negative from (0,4). The local min is at (0,15), and the local max is at (4,35.7). The second derivative is 6x-12, and the concavity is down from (-∞,2) and up from (2,∞). The point of inflection is at (2, 15).
  • #1
hectorubie
4
0

Homework Statement


hey guys. please help me out i tried the problem but for some reason i feel like its wrong. thank you in advance.

Homework Equations


given f(x)=x^3-6x^2+15
a)find the intervals of inc/dec
b)find the local max/min if any
c)find intervals of concavity
d)find the inflection points if any
e)use the information to sketch the graph

The Attempt at a Solution



f'(x)=3x^2-32x
=x(3x-32)
x=0 x=32/3 (critical points)

from -infinity to 0 the function is positive
from 0 to the function is negative
from 32/3 to infinity the function is positive

f(0) = 15 (0,15) = local min
f(32/5) = 35.7 = local max

f''(x)=6x-32
=6x-32=0
=16/3

from -infinity to 16/3 the function is concave down
from 16/3 to infinity the function is concave up

point of inflection (16/3, 15)

cant show the graph here
 
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  • #2


hectorubie said:

Homework Statement


hey guys. please help me out i tried the problem but for some reason i feel like its wrong. thank you in advance.

Homework Equations


given f(x)=x^3-6x^2+15
a)find the intervals of inc/dec
b)find the local max/min if any
c)find intervals of concavity
d)find the inflection points if any
e)use the information to sketch the graph

The Attempt at a Solution



f'(x)=3x^2-32x
=x(3x-32)
x=0 x=32/3 (critical points)

from -infinity to 0 the function is positive
from 0 to the function is negative
from 32/3 to infinity the function is positive

f(0) = 15 (0,15) = local min
f(32/5) = 35.7 = local max

f''(x)=6x-32
=6x-32=0
=16/3

from -infinity to 16/3 the function is concave down
from 16/3 to infinity the function is concave up

point of inflection (16/3, 15)

cant show the graph here

Your derivative is wrong. Try and fix that and try again. And whether the derivative is positive or negative tells you whether the function is increasing or decreasing, yes?
 
Last edited:
  • #3


Your derivative is wrong. Try and fix that and try again. And whether the derivative is positive or negative tells you whether the function is increasing or decreasing, yes?

the first derivative is 3x^2-32x
i factored it to x(3x-32). is that wrong?
 
  • #4


hectorubie said:

Homework Statement


hey guys. please help me out i tried the problem but for some reason i feel like its wrong. thank you in advance.


Homework Equations


given f(x)=x^3-6x^2+15
a)find the intervals of inc/dec
b)find the local max/min if any
c)find intervals of concavity
d)find the inflection points if any
e)use the information to sketch the graph


The Attempt at a Solution



f'(x)=3x^2-32x
=x(3x-32)
x=0 x=32/3 (critical points)

from -infinity to 0 the function is positive
from 0 to the function is negative
from 32/3 to infinity the function is positive

f(0) = 15 (0,15) = local min
f(32/5) = 35.7 = local max

f''(x)=6x-32
=6x-32=0
=16/3

from -infinity to 16/3 the function is concave down
from 16/3 to infinity the function is concave up

point of inflection (16/3, 15)

cant show the graph here

The function f(x) =x^3-6x^2+15 is NOT positive on (-∞,0): when x is large and negative the x^3 term is swamps all the others and is < 0, so f(x) < 0.
 
  • #5


Ray Vickson said:
The function f(x) =x^3-6x^2+15 is NOT positive on (-∞,0): when x is large and negative the x^3 term is swamps all the others and is < 0, so f(x) < 0.

i see where i did wrong there. thanks
 
  • #6


i see where you say the deriv was wrong...
the derive is 3x^2-12x :smile: sorry i didnt notice that i did that
 

1. What does "f(x)" mean in the equation?

"f(x)" is a mathematical notation that represents a function. In this equation, it means that we are looking for the value of the function at a specific input value, denoted by "x".

2. How do I solve the equation?

To solve the equation, we need to find the values of "x" that make the equation true. This can be done by using algebraic methods such as factoring, completing the square, or using the quadratic formula.

3. What is the degree of the polynomial in this equation?

The degree of a polynomial is the highest power of the variable in the equation. In this case, the highest power of "x" is 3, so the degree of this polynomial is 3.

4. How many solutions does this equation have?

Since this is a cubic equation, it can have up to 3 solutions. However, it is possible for some or all of the solutions to be complex numbers.

5. Can I use a graphing calculator to solve this equation?

Yes, a graphing calculator can be a useful tool to find the solutions to this equation. By graphing the equation and using the "zero" or "root" function, we can find the x-values where the graph intersects the x-axis, which correspond to the solutions of the equation.

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