Question

HELP (pic inserted)

Answer 1

the range is all the possible y values from - inf, inf. Look for intervals where the function ceases to exist.

Answer 2

try finding the max

Answer 3

can i just insert equation incalculator

Answer 4

you could graph it?

Answer 5

Or what @chris00 said, you can find the relative max by finding the derivative and setting it equal to 0. then find your x-values (crit point) and these will tell you terminals of increase and decrease for the range AND domain.

Answer 6

gd explanation!

Answer 7

um i dont understand

Answer 8

What do you get when you take the derivative?

Answer 9

im not good at math btw and deritivate of what?

Answer 10

Or better yet,do you know how to find the derivative?

Answer 11

no

Answer 12

Of \(\large f(x) =-6(x-1)^2 +1\)

Answer 13

Ok, so i'm going to show you a quick method of finding it. You can watch youtube videos on finding derivatves of the function to further help you solve the rest of your problems.

Answer 14

ok

Answer 15

You have \[\large f(x)-6(x-1)^2 +1\]\[\large \frac{d}{dx} [f(x)= -6(x-1)^2 +1\]
Apply chain rule to \((x-1)^2\)\[\large f'(x) = -6\frac{d}{dx}[(x-1)^2 +1]\]\[\large -6(2)(x-1)(1)+0=0\] Simplify \(\large -12(x-1)=0\) solve for x.

Answer 16

If you have any further questions about how to find this derivative, @chris00 can assist you :) Good luck with your homework!

And checkout these videos on derivatives https://www.khanacademy.org/math/calculus/differential-calculus/derivative_intro/v/calculus--derivatives-1

Answer 17

how do i solve for x?

Answer 18

this is for college btw -_-

Answer 19

and its kicking my retrice

Answer 20

@Jhannybean how am i suppose to solve for x? im confused :S

Answer 21

@whpalmer4 can you pleasee help me??

Answer 22

easy,

\(\large -12(x-1)=0 \)

Answer 23

use factor theorem :-

x-1 = 0

add 1 both sides

x - 1 = 0
+1 +1

x = 1

Answer 24

next, substitute this value of x in f(x), you would get the max value for range

Answer 25

\(\large f(x) = -6(x-1)^2 +1\)

put x = 1

\(\large f(x) = -6(1-1)^2 +1\)

\(\large = ? \)

Answer 26

f(x)=−6(x−1)^2 +1
now it is always that for x is real,,(x-a)^2,(a is real no.) is always greater than or equal to zero...
so here f(x)is lesser than or equal to 1...(for x belongs to all real values)
so the upper limit is 1...and rest for all x,f(x) is lesser than one...so
for all x belongs to real no. we have f(x) belongs to (-infinity to 1]...

Answer 27

Let f(x)=y
\[y=-6\left( x-1 \right)^{2}+1=-6\left( x ^{2}-2x+1 \right)^{2}+1=-6x ^{2}+12x-6+1\]
\[6x ^{2}-12x+\left( 5+y \right)=0\]
\[x=\frac{ 12\pm \sqrt{12^{2}-4\left( -6 \right)\left( 5+y \right)} }{2*-6 }\]
for x to be real discriminant ge 0
144+120+24y>=0
\[24y \ge -264\]
\[y \ge \frac{ -264 }{ 24 },y \ge -11\]

\[range is [-11,\infty )\]

Answer 28

correction

Answer 29

\[x=\frac{ 12\pm \sqrt{12^{2}-4*6*\left( 5+y \right)} }{2*6 }\]
discriminant=144-120-24y=24-24y
\[24-24y \ge 0 ,-24y \ge -24,y \le \frac{ -24 }{-24 } or y \le 1\]
\[Range is ( -\infty,1]\]