Finding slope of f(x)=3x²-1 at (3,8) and instantaneous rate of change on y=x²-4

[VBoy336]

can you guys help me how to do this problem?

f(x) = 3x^2 - 1 (3,8)
Find slope and tangent line,

I have answer 6, is that right? and can you guys show the step to get the answer, thanks =) (tangent line i already know how to do)

oh and this one,

y = x^2 - 4
Find instantaneous rate of change [3,5]

(if there was one point, i could do it, but i don't know how to do with 2 point,) can you show steps too,


thanks a lot

 

[Rozenwyn]

For the first one, you find the derivative first.
$$f(x) = 3x^2-1 \ \longrightarrow \ f'(x) = 6x$$ then you substitude for x. That doesn't give 6. Now that you have the slope you can find the tangent line.

For the second line, I think that would be the gradient of the secant line through those points.

 

[VBoy336]

for this one,
y = x^2 - 4
Find instantaneous rate of change [3,5]

i don't really get what you are saying, sorry,

so how do I solve it

urg

 

[Rozenwyn]

for this one,
y = x^2 - 4
Find instantaneous rate of change [3,5]

i don't really get what you are saying, sorry,

so how do I solve it

urg

@3, y = 5
@5, y = 11

Now you have 2 points, P1(3,5) & P2(5,11).
What is the gradient of the line that passes through these points ?

 

[VBoy336]

@5 , do you mean 21 ?

and i don't think i learn gradient yet,

in class, when we finding instantaneous rate of change, we use limit x approaching a number, then solve it,

is there another way to find the instantaneous rate of change in this problem?

 

[Rozenwyn]

yeah i mean 21, lol sry. Gradient = slope

 

[VBoy336]

so the instant.. rate of change is 8 ?

aha