Question

Im so confused :(
f(x)=(x^2+4)(3-2x)^2
Solve the inequality for a) f(x)>0 b) f(x)≥0 c) f(x)<0 d) f(x)≤0

Im so confused :(
f(x)=(x^2+4)(3-2x)^2
Solve the inequality for a) f(x)>0 b) f(x)≥0 c) f(x)<0 d) f(x)≤0

@Mathematics

Answer 1

ok first term is always positive because
\[x^2\geq0\] and so
\[x^2+4\geq 4\]

Answer 2

so you can safely ignore that term completely

Answer 3

\[(3-2x)^2>0\] for all x except
\[x=\frac{3}{2}\] in which case you get 0

Answer 4

ignore what term and how do u know it is ≥ and not >?

Answer 5

you can ignore
\[x^2+4\] because it is always greater than or equal to 4, so it is always greater than 0

Answer 6

ok

Answer 7

so the whole thing
\[(x^2+4)(3-2x)^2\] is always greater than or equal to zero, and it is always greater then zero unless x = 3/2
now lets answer each question in turn.

Answer 8

\[a) f(x)>0\]
all x except 3/2

Answer 9

so x cannot equal 3/2 correct?

Answer 10

for that one yes, because if x = 3/2 you get 0

Answer 11

\[b) f(x)≥0 \] true for all x

Answer 12

\[f(x)<0\] never true

Answer 13

\[d) f(x)≤0 \] only true if x = 3/2

Answer 14

so d would be written as (-inf, 3/2]?