Question

How to Evaluate the function at the indicated values.

f(x)= 2x^2+5x-3;

a. f(h+1)
b. f(2x^2 )
c. f(x^2-3)
d. f(x+h)
e. (f(x+h)-f(x))/h,h≠0

please help :)

Answer 1

help ...........

Answer 2

instead of x put h+1 and so on

Answer 3

f(x)= 2x^2+5x-3

Replace all x's with whatever is in the f(...)

eg.

f(h+1) = 2(h+1)^2 + 5(h + 1) - 3

Answer 4

^expand/simplify that

Answer 5

f(x)= 2x^2+5x-3 =(x+3)(2x-1)

Answer 6

^you don't need to factor it...

Answer 7

Simplify the last one. This is the next one, again just replace all x's with the ... in f(...)

\[\Large f(2x^2 ) = 2(2x^2)^2 + 5(2x^2) - 3\]

Answer 8

f(2x^2)=2(2x^2)2+5(2x^2)−3 = 14x^2-3

Answer 9

You're missing a bunch of things... do it step by step \[\Large f(2x^2 ) = 2(2x^2)^2 + 5(2x^2) - 3= \] \[\Large 2(4x^4) + 10x^2 - 3 =\] \[\Large 8x^4 + 10x^2 - 3\]

Answer 10

ok

Answer 11

how about in letter C.f(x^2-3) ??

Answer 12

Try it the same way as the two above... replace all the x's in the equation with (x^2-3), then show us what it looks like immediately after replacing all x's with that.

Answer 13

You're missing something from here...

f(x)= 2x^2+5x-3

Answer 14

f(x^2-3)=2(x^2-3)+5(x^2-3)-3=

Answer 15

ok

Answer 16

Do you see what's missing?

Answer 17

the ^2

Answer 18

f(x^2-3)=2(x^2-3)^2+5(x^2-3)-3=

Answer 19

Correct. Can you simplify it?

Answer 20

And not all at once... step by step if you need to.

Answer 21

7x^2-24 ???

Answer 22

sorry i did again..

Answer 23

Do it step by step, you're missing a lot of terms.

\[\Large 2(x^2-3)(x^2-3)+5(x^2-3)-3=\]start here

Answer 24

2(x^2-3)(x^2-3)+5(x^2-3)-3
2(2x^4-3)+5x^2-3-3
4x^4-3+5x^2

Answer 25

\[\Large 2(x^2-3)(x^2-3)+5(x^2-3)-3=\]
\[\Large 2(x^4-3x^2-3x^2 +9)+5x^2-15-3=\]

Answer 26

"A warm Welcome to OpenStudy. I can guide regarding this useful site; ask your doubts from me, for it you can message me. Please use the chat for off topic questions. And remember to give the helper a medal, by clicking on "Best Answer". We follow a code of conduct, ( http://openstudy.com/code-of-conduct ). Please take a moment to read it."

Answer 27

2x^8-6x^4-6x^4+18 + 5x^2-12

2x^8+18 + 5x^2-12

Answer 28

Wait, where did x^8 come from...? the powers on x don't change when you multiply them by a constant.

Answer 29

hehehe

Answer 30

\[\Large 2(x^4-3x^2-3x^2 +9)+5x^2-15-3= \]
\[\Large 2(x^4-6x^2 +9)+5x^2-18=\]
\[\Large 2x^4-12x^2 +18+5x^2-18=\]

Answer 31

-10x^2+18 + 5x^2-18

Answer 32

You can't combine unlike terms - x^4 and x^2 are unlike terms. The 18 and -18 will cancel off... now just finish

\[\Large 2x^4-12x^2 +18+5x^2-18=\]
\[\Large 2x^4-12x^2 +5x^2=\]now just finish

Answer 33

-10x^2+5x^2

Answer 34

You still can't combine the x^4 and x^2 terms - 2x^4 will stay as 2x^4 unless there's other x^4 terms.

Answer 35

Combine the -12x^2 and +5x^2... those ARE like terms.

Answer 36

2x^4-7

Answer 37

The x^2 can't disappear... it'll still be -7x^2 at the end there.

Answer 38

2x^4-7x^2

Answer 39

Good :)

Answer 40

@ganeshie8 can you help me here ???

Answer 41

i don't know how to solve letter E...

Answer 42

please help me to answer it .. :D

Answer 43

e. (f(x+h)-f(x))/h,h≠0

Answer 44

yeah

Answer 45

this one ?
did u solve letter D ? :)

Answer 46

cuz, we can use its result here

Answer 47

not yet.. but i need to answer letter D first :)

Answer 48

E rather

Answer 49

okiee lets do D first,
then we can use its result in E

Answer 50

uhh okey.. =)

Answer 51

\(\large f(x)= 2x^2+5x-3 \)

\(\large f(x+h) = ?\)

Answer 52

simply replace \(x\) wid \(x+h\)

Answer 53

f(x+h)=2(x+h)^2+5(x+h)-3

Answer 54

Yes !

\(\large f(x)= 2x^2+5x-3 \)

\(\large f(x+h) = 2(x+h)^2+5(x+h)-3\)

Answer 55

simplify,

use te formula
\((a+b)^2=a^2+2ab+b^2\)

Answer 56

wat do u get ?

Answer 57

IDK.

Answer 58

how??

Answer 59

\(\large f(x)= 2x^2+5x-3 \)

\(\large f(x+h) = 2(x+h)^2+5(x+h)-3\)

\(\large f(x+h) = 2(x^2+2xh + h^2)+5x+5h-3\)

Answer 60

see if u can digest above 3rd step :)

Answer 61

and this,

\(\large f(x)= 2x^2+5x-3 \)

\(\large f(x+h) = 2(x+h)^2+5(x+h)-3\)

\(\large f(x+h) = 2(x^2+2xh + h^2)+5x+5h-3\)

\(\large f(x+h) = 2x^2+4xh + 2h^2+5x+5h-3\)

Answer 62

ive just multiplied the 2 inside the stuff in brackets...

Answer 63

yeah..

Answer 64

combine the like terms

Answer 65

you can leave it like that,
or, if u wanto impress ur teacher, just put the function in decreasing degree

Answer 66

uhh ok i dont need to impress her..

Answer 67

hahahah

Answer 68

so where done in letter D

Answer 69

like this :-

\(\large f(x)= 2x^2+5x-3 \)

\(\large f(x+h) = 2(x+h)^2+5(x+h)-3\)

\(\large f(x+h) = 2(x^2+2xh + h^2)+5x+5h-3\)

\(\large f(x+h) = 2x^2+4xh + 2h^2+5x+5h-3\)

\(\large f(x+h) = 2x^2+x(4h+5) + 2h^2+5h-3\)

Answer 70

WoW thanks @ganeshie8

Answer 71

yes, we're done wid D

Answer 72

next look at E,

Answer 73

yep

Answer 74

E : \(\large \frac{\color{red}{f(x+h)} - f(x)}{h}\)

Answer 75

we worked the thing in red just now

Answer 76

just plug its value there,
and simplify if possible

Answer 77

uhh

Answer 78

\(\huge \frac{\color{red}{f(x+h)} - f(x)}{h}\)

from D, we knw f(x+h) already,


\(\huge \frac{\color{red}{2x^2+4xh + 2h^2+5x+5h-3} - f(x)}{h}\)

Answer 79

fine so far ?

Answer 80

yep..

Answer 81

plugin f(x) value also
and simplify watever u can

Answer 82

ok

Answer 83

so that it..

Answer 84

\(\huge \frac{\color{red}{f(x+h)} - f(x)}{h}\)

from D, we knw f(x+h) already,


\(\huge \frac{\color{red}{2x^2+4xh + 2h^2+5x+5h-3} - f(x)}{h}\)

plugin f(x) value

\(\huge \frac{\color{red}{2x^2+4xh + 2h^2+5x+5h-3} - (2x^2+5x-3)}{h}\)

Answer 85

where did we get the f(x) = (2x^2+5x-3)??

Answer 86

look at ur question,
f(x) = (2x^2+5x-3) is the first sentense in ur question :)

Answer 87

uhh i know na its from the value

Answer 88

ok

Answer 89

it simplifies nicely
give it a try...

Answer 90

ok we will just simplifying it

Answer 91

2x^2+x(4h+5)+wh^2+5h-3-(2x^2+5x-3) / h

Answer 92

??? @ganeshie8

Answer 93

do you think my answer is already correct??

Answer 94

\(\huge \frac{\color{red}{f(x+h)} - f(x)}{h}\)

from D, we knw f(x+h) already,


\(\huge \frac{\color{red}{2x^2+4xh + 2h^2+5x+5h-3} - f(x)}{h}\)

plugin f(x) value

\(\huge \frac{\color{red}{2x^2+4xh + 2h^2+5x+5h-3} - (2x^2+5x-3)}{h}\)

\(\huge \frac{\color{red}{2x^2+4xh + 2h^2+5x+5h-3} - 2x^2-5x+3}{h}\)

\(\huge \frac{\color{red}{4xh + 2h^2+5h}}{h}\)

Answer 95

you can still simplify it...

Answer 96

uhh