Asked by Colin
Okay i have a few function questions im gonna post. Im completely confused with these so any help would be great.
Minimum or maximum value of:
f(x)=2x^2+4x+1
and
f(x)=x^2-3x+2
then
if f(x)=x^2+9 and g(x)=x-4,
find (g*f)(-3)
next is
if f(x)=x^3+x^2-x-1 and g(x)=x+1,
find (f/g)(n)
and the possible solutions to the problem are
A) n^2-x-1
B) (n^2+x-1/n+1)
C) n+1
D) n^2-1
and last, find the zeroes of
f(x)=2x^2-10
thanks for any help!
Minimum or maximum value of:
f(x)=2x^2+4x+1
and
f(x)=x^2-3x+2
then
if f(x)=x^2+9 and g(x)=x-4,
find (g*f)(-3)
next is
if f(x)=x^3+x^2-x-1 and g(x)=x+1,
find (f/g)(n)
and the possible solutions to the problem are
A) n^2-x-1
B) (n^2+x-1/n+1)
C) n+1
D) n^2-1
and last, find the zeroes of
f(x)=2x^2-10
thanks for any help!
Answers
Answered by
drwls
<< find the zeroes of
f(x)= 2x^2-10 >>
f(x) = 2(x^2 -5)
= 2(x + sqrt 5)(x - sqrt 5)
When either of the x-containing factors is zero, f(x) is zero.
<<Minimum or maximum value of:
f(x)=2x^2+4x+1
and .. >>
Since you aren't taking calculus and presumably do not know about deriviatives, I suggest you use the method of "completing the square"
f(x)= 2(x^2 + 2x) + 1
= 2 (x^2 + 2x + 1) -1
= 2(x+1)^2 -1
It should be evident that this will be have a minimum when x+1 = 0. That tells you what x is there.
f(x)= 2x^2-10 >>
f(x) = 2(x^2 -5)
= 2(x + sqrt 5)(x - sqrt 5)
When either of the x-containing factors is zero, f(x) is zero.
<<Minimum or maximum value of:
f(x)=2x^2+4x+1
and .. >>
Since you aren't taking calculus and presumably do not know about deriviatives, I suggest you use the method of "completing the square"
f(x)= 2(x^2 + 2x) + 1
= 2 (x^2 + 2x + 1) -1
= 2(x+1)^2 -1
It should be evident that this will be have a minimum when x+1 = 0. That tells you what x is there.
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