Functions and Function Notation?

Question

Let f(x)=x^3-7x^2+5x
Find all values of x for which f(x)= -1
(Kindly show all solutions. I'm still a fourth year highschool student so I don't know.)

Answer 1

f(x)=x^3-7x^2+5x && f(x)=-1
-1=x^3-7x^2+5x
x^3-7x^2+5x +1 = 0
(x-1) (x^2-6x-1) = 0
(x-1)(x-3-10^(1/2))(x-3+10^(1/2))=0
so x=1 or x=3-10^(1/2) or x=3+10^(1/2)

Answer 2

Hi,

well, if you have: f(x) = x^3 - 7x^2 - 5x, you are looking for 3 differents values of (x), because this is a third grade equation , where f(x) is -1
so, x^3 - 7x^2 - 5x +1 = 0,
now we can put (-7x^2) like (-6x^2- x^2)
(x^3 - x^2) + ( -6x^2 -5x +1 ) = 0
x^2(x-1) = 6x^2 + 5x -1,
x^2(x-1) = (6x + 1)(x-1), now you already that if (x-1) = 0, x=1
so, x=1 is your first answer.
Now, x^2 = 6x +1
x^2 - 6x -1 = 0
now working with (x^2 - 6x - 1), and remembering the quadratic formula:

X = {-b (+)or(-) (sqrt)[ b^2 - (4)(a)(c)] } / 2(a)

now in our expression: (x^2 - 6x - 1) ,
a= 1, b= -6, and c=(-1)

so, X= {(-6) (+)or(-) sqrt[ (-6)^2 - 4(1)(-1) ]} / 2(1)

X= {6 (+) or (-) sqrt[ 40 ]} / 2
Therefore:
X= (3 + sqrt10), or (3 - sqrt10)
Thus, your three answers are:
X=1, X=(3 + sqrt10), or (3 - sqrt10)

I hope it helped!

Answer 3

There will be, at most, 3 solutions, because the highest degree of this polynomial is 3.

You can solve it by setting it equal to -1, because thats what your trying to find, then get it equal to zero, and solve for x.
x^3 - 7x^2 + 5x = -1
x^3 - 7x^2 + 5x + 1 = 0

When you factor it you get:
(x - 1)(x^2 - 6x - 1) = 0

Now, set each term equal to zero, and solve.
x - 1 = 0
x = 1

x^2 - 6x - 1 = 0
(using the quadratic equation, you get:)
x = -(sqrt(10) - 3) or x = sqrt(10) + 3.

Thus, the values for x that would make f(x) = -1 are: 1, -(sqrt(10) - 3), and sqrt(10) + 3. Or, a more approximated value for them: 1, -0.162278, and 6.16228

Answer 4

to be a function, each x value has to have exactly one y value. Consider x = 1. Then 3x+2 = 3(1)+2=5. Now, how many y values are greater than or equal to this? Infinitely, for example 6, 7, 8..... all are greater than 5.

Answer 5

x³ - 7x² + 5x = - 1
x³ - 7x² + 5x + 1 = 0
Use synthetic division (or otherwise) to show that x - 1 is a factor (* used to spaceout typing):-
1|1****-7****5*****1
*|******1*** - 6** - 1
-------------------------
*1*****-6***-1*****0 = R
------------------------
This shows that, because R = 0, x - 1 is a factor.
Other factor is x² - 6x - 1
f(x) = (x - 1).(x² - 6x - 1) = 0

x = 1 , x = [ 6 ±√(40) ] / 2
x = 1 , x = [ 6 ± 2√10 ] / 2
x = 1 , x = 3 ± √10

Answer 6

f(x)=x^3-7x^2+5x f(x)=-1
-1=x^3-7x^2+5x
x^3-7x^2+5x +1 = 0
(x-1) (x^2-6x-1) = 0
(x-1)(x-3-10^(1/2))(x-3+10^(1/...
so x=1 or x=3-10^(1/2) or x=3+10^(1/2)

Answer 7

so we have a function & you say f(x)=-1 so you should number -1 instead of all x
so we have f(-1)=(-1)^3-7(-1)^2+5x
f(-1)=-1-7-5= -13

Answer 8

x^3-7x^2+5x=-1
p(x) = x^3-7x^2+5x+1=0
p(1)=0 ==> p(x)=(x-1)q(x)

q(x) = p(x) /(x-1) = x^2-6x-1
x^3-7x^2+5x+1=(x-1)(x^2-6x-1)
x^3-7x^2+5x+1=0 ==> x-1=0 & x^2-6x+1=0
x1 = (6+sqrt(40))/2 & x2=(6-sqrt(40))/2

Answer 9

(-1)^3 - 7(-1)^2 + 5(-1)

-1 - 7(1) - 5
-1-7-5
-8-5
-13