Sketching functions help please!

Would some please explain a step by step procedure to sketching graphs? Without calculus. I know i have to find the domain, poles and zeroes,(i know how to find the former three), and all its asymptotes.
But Im not quite sure how to tackle the way in which the functions changes...and how to tell how the function behave ( as in which direction its heading towards). When its at its min point etc. Or when the function splits.

Please explain how to do so! It would be greatly appreciated!

Original post by Trufflezz

Would some please explain a step by step procedure to sketching graphs? Without calculus. I know i have to find the domain, poles and zeroes,(i know how to find the former three), and all its asymptotes.
But Im not quite sure how to tackle the way in which the functions changes...and how to tell how the function behave ( as in which direction its heading towards). When its at its min point etc. Or when the function splits.

Please explain how to do so! It would be greatly appreciated!

Solve f(x)=0 to find roots (where graph crosses x axis)

Set x to 0 to find where graph crosses y axis.

Differentiate function, set f'(x)=0 for turning points

Differentiate again f''(x) to find nature of turning points.

If second derivative >0 SP is minimum

If second derivative <0 SP is maximum

If second derivative = 0, point of inflexion.

I know you said no calculus.. So you could complete the square on the function to find the info, however not always possible/easy to complete the square.

Hope that helped.

SP is turning point or stationary point

(edited 9 years ago)

Reply 2

username1732133

15

Original post by Trufflezz

Would some please explain a step by step procedure to sketching graphs? Without calculus. I know i have to find the domain, poles and zeroes,(i know how to find the former three), and all its asymptotes.
But Im not quite sure how to tackle the way in which the functions changes...and how to tell how the function behave ( as in which direction its heading towards). When its at its min point etc. Or when the function splits.

Please explain how to do so! It would be greatly appreciated!

Since you have said that you do not want to use calculus here's a simple option.

For many of your questions it will be helpful to factorise. Having done that you can consider the signs of the factors in various intervals.

Why not chose one of your eleven questions to work through and get some help with?

Reply 3

Trufflezz

OP

Original post by BuryMathsTutor

Since you have said that you do not want to use calculus here's a simple option.

For many of your questions it will be helpful to factorise. Having done that you can consider the signs of the factors in various intervals.

Why not chose one of your eleven questions to work through and get some help with?

can u help me out with question (b) and (f)? Please? Thanks a bunch!

Reply 4

username1732133

15

Original post by Trufflezz

can u help me out with question (b) and (f)? Please? Thanks a bunch!

Yes. Were you able to apply my hints to (b)? What did you get?

Reply 5

Trufflezz

OP

Original post by BuryMathsTutor

Yes. Were you able to apply my hints to (b)? What did you get?

negative in the first interval, positive in the 2nd and 3rd, negative in the fourth?

I plugged in x values in those intervals and plotted them...its that how i was suppose to do it?

or is there a quicker way?

Reply 6

Trufflezz

OP

fpr ppart (f) i dnt even know how to start...its such an ugly function ... :frown:

Reply 7

Hasufel

16

Original post by Trufflezz

fpr ppart (f) i dnt even know how to start...its such an ugly function ... :frown:

factorise the numerator in terms of

\sqrt{2}

.. (then cancel down)

f

simplifies to: (verify this!):

\displaystyle \frac{1}{x-\sqrt{2}}

so it`s not really that ugly...

(edited 9 years ago)

Reply 8

poorform

10

Original post by ApplyYourself

Solve f(x)=0 to find roots (where graph crosses x axis)

Set x to 0 to find where graph crosses y axis.

Differentiate function, set f'(x)=0 for turning points

Differentiate again f''(x) to find nature of turning points.

If second derivative >0 SP is minimum

If second derivative <0 SP is maximum

If second derivative = 0, point of inflexion.

I know you said no calculus.. So you could complete the square on the function to find the info, however not always possible/easy to complete the square.

Hope that helped.

SP is turning point or stationary point

Not necessarily. Consider

f:\mathbb{R}\rightarrow\mathbb{R}

given by

f(x)=x^4

.

Reply 9

ApplyYourself

16

Original post by poorform

Not necessarily. Consider

f:\mathbb{R}\rightarrow\mathbb{R}

given by

f(x)=x^4

.

Yeah, C2 that's what I was told (point of inflexion) ... I blame the books

Reply 10

poorform

10

Original post by ApplyYourself

Yeah, C2 that's what I was told (point of inflexion) ... I blame the books

It can be a point of inflexion but it doesn't have to be as my example shows. To be honest if

\displaystyle f''(x)=0

then you could either use the first derivative test or compare

\displaystyle f''(x)

both above and below

\displaystyle c

to see if the concavity has changed.

Reply 11

Hasufel

16

f) is kinda difficult to do without some form of calculus - limits essentially.

the fuction you end up with is:

\displaystyle \frac{1}{e^{x-\sqrt{2}}}-1

which approaches

y=-1

as

x

approaxhes

\sqrt{2}

from the left - (this is a {global} minimum point)

and approaches

\infty

as

x

approaches

\sqrt{2}

from the right

then what does y approach (easy this) as x approaches +/- infinity?...

(also, there is a point of inflection below x=-1, but without calculus....)

(edited 9 years ago)

Sketching functions help please!

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