Question

if f(x)=4x^3-x^2-2x+1 and g(x)={min {f(t):0<=t<=x for 0<=x<1 and 3-x for 1 11 years ago

Answer 1

Is this calculus?

Answer 2

f(x)=4x^3-x^2-2x+1 and g(x)={min {f(t):0<=t<=x for 0<=x<1 and 3-x for 1<x<=2} then find the value of g(1/4)+g(1/2)+g(5/4)? I'm having trouble seeing what the equations are supposed to look like on here?

Answer 3

2 definitions of g(x) are given !

Answer 4

what are we trying to find?

Answer 5

g(1/4)+g(1/2)+g(5/4)

Answer 6

if f(x)=4x^3-x^2-2x+1 and g(x)={min {f(t):0<=t<=x for 0<=x<1 and 3-x for 1<x<=2} then find the value of g(1/4)+g(1/2)+g(5/4)
\[\Large f(x)=4x^3-x^2-2x+1\]
\[ g(x)=\left\{ \min \left\{ f(t):0 \leq t \leq x \right\} \right\}, 0 \leq x<1\]
\[\Large g(x)=3-x , 1<x \leq 2\]

------------------------------------
We need :
\[\Large g(1/4)+g(1/2)+g(5/4)\]
We have 2 cases in g(x) so first we will decide which will we evaluate.
if our \(\Large x>1 \) OR \(\Large x \leq 2\)
and out of 1/4 , 1/2 and 5/4 only 5/4 satisfies this so..
\[\Large g(\frac{5}{4})=3-\frac{5}{4}=\frac{7}{4}\]
Similarly to evaluate 1/2 and 1/4 you need to refer to the former definition and then add all 3 up!

Answer 7

that is easily seen for g(5/4) how to go for g(1/2) and g(1/4)? how to evaluate the first condition?

Answer 8

@DLS ?

Answer 9

@ganeshie8 ?

Answer 10

@Abhisar ?

Answer 11

@beccaboo333 ?

Answer 12

@CaseyCarns ?

Answer 13

@dan815 ?

Answer 14

@eric_d ?

Answer 15

@Future_UsArmy_MP ?

Answer 16

@hartnn ?
@Hero ?
@it_meh_miranda ?

Answer 17

@JFraser ?
@Koikkara ?
@Luigi0210 ?
@mathslover ?
@nincompoop ?
@oksuz_ ?
@ParthKohli ?
@qdbug @
@RosieF ?
@SithsAndGiggles ?
@Tushara ?

Answer 18

I have no clue sry

Answer 19

ok

Answer 20

anyone?

Answer 21

i haven't done this kind of math in a while, sorry

Answer 22

Allow me to rewrite the given functions so it's a bit clearer to see:
\[f(x)=4x^3-x^2-2x+1\]
\[g(x)=\begin{cases}\min\{f(t):0\le t\le x\}&\text{for }0\le x<1\\3-x&\text{for }1<x\le2\end{cases}\]
You want to find \(g\left(\dfrac{1}{4}\right)+g\left(\dfrac{1}{2}\right)+g\left(\dfrac{5}{4}\right)\).
\[g\left(\frac{1}{4}\right)=\min\left\{f(t):0\le t\le\frac{1}{4} \right\}\]
The minimum of \(f\) over the interval \(\left[0,\dfrac{1}{4}\right]\) can be determined with some calculus, using the test for extrema. The same can be said for \(g\left(\dfrac{1}{2}\right)\).

Answer 23

Hmm, the less than/equal to signs aren't showing up for me...

Answer 24

\[f'(x)=12x^2-2x-2\]
Critical values:
\[\begin{align*}12x^2-2x-2&=0\\
6x^2-x-1&=0\\
(3x+1)(2x-1)&=0\\
x&=-\dfrac{1}{3},~\dfrac{1}{2}
\end{align*}\]
The negative solution can be ignored, since \(g\) isn't defined for negative values of \(x\).

Over the interval \(\left(0,\dfrac{1}{2}\right)\), \(f'(x)\) is negative, so \(f(x)\) is decreasing.
Over \(\left(\dfrac{1}{2},\infty\right)\), \(f'(x)\) is positive, so \(f(x)\) is increasing.

Thus \(f(x)\) has a minimum at \(x=\dfrac{1}{2}\).

The minimum of \(f\) for \(\left[0,\dfrac{1}{4}\right]\) must occur at an endpoint:
\[f(0)=1~~\text{and}~~f\left(\frac{1}{4}\right)=\frac{1}{2}\]
So, \(g\left(\dfrac{1}{4}\right)=\min\left\{f(t):0\le t\le\dfrac{1}{4}\right\}=\dfrac{1}{2}\).

Answer 25

Similarly, \(g\left(\dfrac{1}{2}\right)=\min\left\{f(t):0\le t\le\dfrac{1}{2}\right\}=\dfrac{1}{4}\), since
\[f(0)=1~~\text{and}~~f\left(\frac{1}{2}\right)=\frac{1}{4}\]
To recap,
\[g\left(\dfrac{1}{4}\right)+g\left(\dfrac{1}{2}\right)+g\left(\dfrac{5}{4}\right)=\frac{1}{2}+\frac{1}{4}+\frac{7}{4}=\frac{5}{2}\]
Kudos to @DLS for \(g\left(\dfrac{5}{4}\right)\).