Question

F(x) = -x^2 –x+1

G(x) = -x^2 +x+6

Use the given functions f and g to solve for the following:


(d) Solve f(x) >0

(e) Solve g(x) ≤ 0

(f) Solve f(x) > g(x) F(x) = -x^2 –x+1

G(x) = -x^2 +x+6

Use the given functions f and g to solve for the following:


(d) Solve f(x) >0

(e) Solve g(x) ≤ 0

(f) Solve f(x) > g(x) @Mathematics

Answer 1

first one you are going to have to set it equal zero and use the quadratic formula

Answer 2

then since it is a parabola that opens down, it will be positive between the zeros.
same for the second one

Answer 3

how many answers are there to this ?

Answer 4

not sure exactly what you mean. there will be one interval for the first problem. it contains and infinite number of numbers. lets just do it

Answer 5

is this correct? -x^2-x+1>0
-x^2-x>1
x^2+x>-1
x^2+x+1/4>5/4
(x+1/2)^2>5/4
|x+1/2| > (sqrt 5) /2
1/2(1−5√)<x<12(5√−1)

Answer 6

yes

Answer 7

ok- so am I missing anything to this answer?

Answer 8

well sort of

Answer 9

your answer is correct. but the solution contains a mistake

Answer 10

1/2(1−5√)<x<1/2(5√−1)

Answer 11

the correct answer you have
it is
\[\frac{1-\sqrt{5}}{2}<x<\frac{1+\sqrt{5}}{2}\]

Answer 12

ok. Is that the entire answer? I feel like I am missing something. LOL

Answer 13

that is right. for the first one

Answer 14

ok- guess I am not sure of the next two

Answer 15

there is a mistake in your procedure, but the answer is right. you made two mistakes and they canceled each other

Answer 16

it should be
\[-x^2-x+1>0\]
\[-x^2-x>-1\]
\[x^2+x<1\] etc.

Answer 17

finally to
\[|x+\frac{1}{2}|<\frac{\sqrt{5}}{2}\] an then
\[\frac{-1-\sqrt{5}}{2}<x<\frac{-1+\sqrt{5}}{2}\]

Answer 18

second one is easier. get
\[-x^2+x+6\leq0\]
\[-(x-3)(x+2)\leq 0\]
\[x\leq -2\text { or } x \geq 3\]

Answer 19

third one easiest of all, get
\[-x^2 –x+1>-x^2 +x+6\]
\[-x+1>x+6\]
\[-5>2x\]
\[-\frac{5}{2}>x\]

Answer 20

ok- I was wondering what I did wrong the first time...I had some wierd answer- similar to what you put but a few mistakes, but this makes more sense. Thank you so much.

Answer 21

this is what i had originally...(e)
Since the zeros of the function are x1, 2 = −1 ± √1+24 /2 = −1 ± 5 /2= 2 and -3.

The solution is, ∀ x ∈ (−∞, −3] V [2, +∞). This time the values of the zeros of the function are accepted.

(f)
You have to evaluate f-g>0 since f (x) − g(x) = −2x −5. The solution is ∀ x > −5/2