Derivative Problem: f’(1), f’(2), f’(3), and f’(5)

[SyaharaAden]

Homework Statement

Derivative Problem

Relevant Equations

Derivative

1. How to get f’(1), f’(2), f’(3), and f’(5)
2. How to calculate average speed change of y to changes in x in the interval [0,6]
3. Estimate value of f’+(0) and f’-(6)

pls help me about this graph, i dont know how to read this

 

[BvU]

Hello, and

Here at PF we need something from you before we are allowed to help (see guidelines). So:

What do you know about derivatives ? definition ?

$\$

 

[haruspex]

… and how is an average acceleration defined?

 

[Steve4Physics]

I don't think acceleration is needed. In fact, maybe the questions are not about motion at all. Possibly the intended questions are these:

1. Find $f’(1), f’(2), f’(3)$ and $f’(5)$ (where $f'$ denotes $\frac {dy}{dx}$).

2. Calculate the average rate of change of $y$ with respect to $x$ over the interval from $x=0$ to $x=6$.

3. Estimate the values of $f’_+(0)$ and $f_-'(6)$.

However, I suspect we might not hear from the poster again! (But - if they are reading this - I hope my suspicion is wrong!)

 

[haruspex]

I don't think acceleration is needed. In fact, maybe the questions are not about motion at all.

The OP's reference to speed change leads me to suppose the full question sets the context as motion.
Btw, I do not like the $f'(1)$ part. The diagram is not drawn correctly for that case. The straight line matches the slope of the curve somewhere around $x=\frac 12$, not at $x=1$.

 

[Steve4Physics]

The OP's reference to speed change leads me to suppose the full question sets the context as motion.

The phrase used by the OP is "calculate average speed change of y to changes in x ...".

There is no indication that ‘x’ represents time, so it seems possible (likely?) that the OP has used 'speed' thinking it is interchangeable with the term 'rate of change'. But we may never know!

Btw, I do not like the $f'(1)$ part. The diagram is not drawn correctly for that case. The straight line matches the slope of the curve somewhere around $x=\frac 12$, not at $x=1$.

Yes. I'd draw some of the other tangents slightly differently too.

 

[SyaharaAden]

sorry for typo, yes it should be “rate of change”, my bad when i’m translated the question in my language to english

i’m still confused how to answer 3 of question, is there the link for explanation something like my op

(sorry if my english are bad)

 

[Steve4Physics]

sorry for typo, yes it should be “rate of change”, my bad when i’m translated the question in my language to english

i’m still confused how to answer 3 of question, is there the link for explanation something like my op

(sorry if my english are bad)

Hi @SyaharaAden. The rules here require you to show evidence of your own effort before we give you help. See what @BvU said in Post #2.

So:

a) Tell us what you got for Q1 and Q2 and how you got the answers. This will allow us to check your understanding. (You can't do Q3 unless you fully understand Q1 and Q2).

b) Tell us what you think $f’_+$ and $f’_-$ mean.

 

[SyaharaAden]

a) Tell us what you got for Q1 and Q2 and how you got the answers. This will allow us to check your understanding. (You can't do Q3 unless you fully understand Q1 and Q2)

what my lecture explain when study before is just about what the definition about f’(x), in his explanation :
f’(x) = lim h->0 [(f(x+h) - f(x)]/h
and then :
f’(x) = 1/(2*sqrt[x])

no explanation in my lecture about somethink like read the graph in my op, also the graph type isn’t familiar for me

if i follow the equation, then f’(1) is :
1/(2*sqrt[1])
= 1/2
f’(2) :
1/(2*sqrt[2])
=1/(2*sqrt[2]) * sqrt[2]/sqrt[2]
=sqrt[2]/4
(So basically, i’m not read/analys the graph, i just substitute the value to equation, so what function the graph is?)
is that true for Q1? if yes i will try explain how i find the Q2, if its not true, pls explain me first for Q1

 

[haruspex]

what my lecture explain when study before is just about what the definition about f’(x), in his explanation :
f’(x) = lim h->0 [(f(x+h) - f(x)]/h

Good as an exact definition, but doesn't give you an intuitive grasp. In terms of the graph, it is the slope of the graph at x. The diagram has usefully provided tangent lines at the points you are asked about in the first part, and a tangent at a point of the graph is a straight line with the same slope there.

and then :
f’(x) = 1/(2*sqrt[x])

That is not generally true. This must have been in regard to a specific function, like $f(x)=\sqrt x$. It is not relevant to this question.

 

[Steve4Physics]

@SyaharaAden , in addition to what @haruspex said in Post #10:

1. What is the gradient of the sloping straight line passing through the point (2, 3) on your graph?

2. Is this line a tangent to the curve at the point (2,3)?

 

[FactChecker]

Those tangent lines are there so that you can visually estimate the derivative at those x-values. Do you know the relationship between a derivative and the slope of a tangent line on the graph? For each tangent line, you can pick some points on it that will give you an easy slope calculation.