Question

Let f(x) = x^3 + bx, where b is an element of the real numbers. Find all value(s) of b such that f`(x)=0 for all real values of x.

Answer 1

f'(x) = 0 means:
f'(x) = 3x^2+b = 0

Answer 2

3x^2 + b = 0
3x^2 = -b
So now, since the left side can only be a positive number.

Answer 3

And we're only dealing with reals in here, b can range from -infinity to 0, or:
(-inf, 0]

Answer 4

basically b has to be non-positive.

Answer 5

hm..i don't see why

Answer 6

because 3*x^2 cannot equal to a negative number

Answer 7

3 = positive
x^2 = positive (any number squared is positive)
positive*positive = positive
3*(x^2) = -b

Answer 8

thus -b must be positive, or b must be negative.. here let me use an example.

Answer 9

Suppose b = 1
3x^2 = -1
x^2 = -1/3
but we are only dealing with reals, not imaginary numbers. That's why b cannot be positive.

Answer 10

Now suppose b = 0
3*x^2 = 0
x^2 = 0, x = 0. So 0 works.

Answer 11

Now suppose b = -1
3*x^2 = -(-1)
3x^2 = 1
x^2 = 1/3
x = +/- sqrt(1/3)
See. This only works with 0 or negative numbers.

Answer 12

okay. i am following what you're saying. just have to form it into words. my professor is really strict with forming our proofs.

Answer 13

WAIT a second lol

Answer 14

what'd i do? haha

Answer 15

i forgot, it didn't just say exists, it said = 0.

Answer 16

ok...

Answer 17

oh nevermind i was right, i did take that into account.

Answer 18

did it matter that you solved for b though when the question asked for all real values of x? is it referring to once you've solved the equation completely?

Answer 19

x is real means that that it's square cannot be negative.

Answer 20

that's why I knew that b had to be negative.

Answer 21

Okay let's start over:
f'(x) = 3x^2 + b = 0
3x^2 + b = 0
3x^2 = -b
x^2 = -b/3
So far so good, right?

Answer 22

x is a real number, therefore \[x^2 \ge 0\]
So:
\[-b/3 \ge 0\]

Answer 23

okay...now it's in terms of x.

Answer 24

Multiplying both sides of the equation by 3 gives:
\[-b \ge 0\]
Now multiply both sides by -1, but remember when multiplying by a negative sign, u reverse the inequality, so you end up with:
\[b \le 0\]

Answer 25

good?

Answer 26

good. back to my last question...i wrote a proof out. would you let me know if it makes any sense

Answer 27

kk hold on