Question

f(x)=e^3,g(x)=2^x,h(x)=5 then
fogofogohogofohogof(x)= ?

Answer 1

can't anybody solve this simple question?

Answer 2

Provided that you actually do mean \(f(x)=e^3\) (a constant), the given function
\[f\circ \cdots(x)=f(x)\]

Answer 3

think again sithsandgiggles

Answer 4

\((\large f\circ l)x=e^3\), since \(f\) is not a function of x.

Here , \(\large l=g\circ f\circ g\circ h\circ g\circ f\circ h\circ g\circ f\), so regardless of \(\large l\), \(\large(f\circ l)x=f(x)=e^3\\\)

What are your ideas?

Answer 5

^^exactly.

Answer 6

Thats a lot of work. I skip questions like this on real exams on purpose.

Answer 7

it is just \(e^3\)

Answer 8

oh, i am late
what @ybarrap said

Answer 9

look @ all three f'ns, f(x) & g(x) both are contant f'ns and hence both of them don't have any
variable term as out put so after one of them apeears in gof... the funtion becomes not defined after it is provided input to g(x) hence the ans is not defined

Answer 10

Let's simplify this question a bit.

Suppose \(f(x)=1\) and \(g(x)=2x\). Then \((f\circ g)(x)=f(g(x))=f(2x)=1\). It doesn't matter what the value of \(x\) is; for any input you give for \(f\), you will always get 1 as an output.

Answer 11

k friend i agree that for any value of x f(x) =e^3 but there should be a 'x' so that the function gets a number as input
u can't say f(ball)=e^3 as ball is not a number

Answer 12

Well, neither is \(x\). It's a letter, but it happens to take on any number we like because *that's how we define it*. It's an element of a set of numbers.

You could say the same for "ball." If we take ball to mean some variable, then \(f(ball)=1\).

Answer 13

if we take ball to mean some variable but first it should be defined to be number

Answer 14

\[\Large f(x)=e^3, \qquad g(x)=2^x,\qquad h(x)=5\]
\[\Large f(g(f(g(h(g(f(h(g(\color{royalblue}{f(x)})))))))))\]
Just work from the inside out, starting with this blue part.

\[\Large =f(g(f(g(h(g(f(h(g(\color{royalblue}{e^3})))))))))\]\[\Large =f(g(f(g(h(g(f(h(\color{orangered}{g(e^3)}))))))))\]\[\Large =f(g(f(g(h(g(f(h(\color{orangered}{2^{e^3}}))))))))\]\[\Large =f(g(f(g(h(g(f(\color{green}{h(2^{e^3})})))))))\]\[\Large =f(g(f(g(h(g(f(\color{green}{5})))))))\]

Answer 15

\[\Large =f(g(f(g(h(g(\color{#CC0033}{f(5)}))))))\]\[\Large =f(g(f(g(h(g(\color{#CC0033}{e^3}))))))\]\[\Large =f(g(f(g(h(\color{#3399AA}{g(e^3)})))))\]\[\Large =f(g(f(g(h(\color{#3399AA}{2^{e^3}})))))\]

Answer 16

\[\Large =f(g(f(g(\color{#FF11AA}{h(2^{e^3})}))))\]\[\Large =f(g(f(g(\color{#FF11AA}{5}))))\]\[\Large =f(g(f(\color{darkorchid}{g(5)}))))\]\[\Large =f(g(f(\color{darkorchid}{2^{e^5}}))))\]

Answer 17

\[\Large =f(g(\color{orangered}{f(2^{e^5})})))\]\[\Large =f(g(\color{orangered}{e^3})))\]\[\Large =f(\color{blue}{g(e^3)})\]\[\Large =f(\color{blue}{2^{e^3}})\]

Answer 18

\[\Large =\color{purple}{f(2^{e^3})} \qquad=\qquad e^3 \qquad=\qquad f(x)\]

Answer 19

I know these guys already explained this one pretty well for you, but that looked like fun to go through lol.

@coolgeek what part are you confused on?

Answer 20

The inner inner inner most function is f(x), which is defined to be a constant for ALL values of x.

We don't have to specific which x we're using for this portion, it cannot be undefined for any value of x.

Answer 21

oh boy :D yer typing up a storm there huh? lol

Answer 22

@zepdrix i appreciate u r work but i went through a book where author had written this ans 4 the question

Answer 23

Maybe a simple example will help.

\[\Large y=y(x)\]\[\Large y=1\]For what values of x is the function y undefined?

There are no values for which y is undefined. It is defined to be 1 for all x.
We can't claim that y(x) is undefined because we didn't specify a point.
It is 1 regardless of the point. It can't be undefined!! :O

Books makes mistakes sometimes D: I guess...
Maybe the question was written down incorrectly? <:o