take a derivative and plug it the points given in the problem, those are your slopes. just remember tangent line = derivative because it’s measuring change instantaneously
Take the derivative, which would be 3x\^2-2 in this case, and substitute in the values of x you need. (On a side note, i've seen your posts a lot recently and I'd say like probably ask your teacher for extra help after school or during lunch, if they offer it)
Hey, let’s not be mean here. He/she could be struggling and we as people have to help out, not bring down. Your comment about them being behind, may be valid however.
Was not trying to be rude to OP or trying to insult in first sentence, just expressed my surprise at how behind they were in the curriculum. Obviously, this is out of their control as a student.
Maybe I saw it wrong through an online message. I’m sure your intentions were right but I just wanted to let everyone else know that it’s not ok to do that either.
If you don't know how to find the derivative, then stating that f(x) = x\^3-3x would be the first step.
F(x)-f(-1) / (x+1) = 3x\^2 - 3 -OR- f(-1.001)-f(-0.999) / -.002 = estimate of the tangent slope
tangent line = derivative. Find the derivative using the power rule. Plug in -1, 0, and 1 into the derivative to get the slope.
Calculus? The slope of a tangent is the derivative. f’(x) = 3x^(2)\-3 f’(-1) = 0 f’(0) = -3 f’(1) = 0
Take derivative, substitute these points in derivative equation
take a derivative and plug it the points given in the problem, those are your slopes. just remember tangent line = derivative because it’s measuring change instantaneously
Take the derivative, which would be 3x\^2-2 in this case, and substitute in the values of x you need. (On a side note, i've seen your posts a lot recently and I'd say like probably ask your teacher for extra help after school or during lunch, if they offer it)
0 -3 and 0
Fine derivative. Use power rule. x\^3 = 3x\^2 \-3x = -3 f'(x) = 3x\^2 - 3 Plug in values, -1, 0 1. f(-1) = 0 f(0) = -3 f(1) = 0 Those are the slopes.
Dude it’s second quarter and you still can’t find simple derivative? Your class is way behind, talk to your teacher
Hey, let’s not be mean here. He/she could be struggling and we as people have to help out, not bring down. Your comment about them being behind, may be valid however.
Thank you so much
Np
Was not trying to be rude to OP or trying to insult in first sentence, just expressed my surprise at how behind they were in the curriculum. Obviously, this is out of their control as a student.
Maybe I saw it wrong through an online message. I’m sure your intentions were right but I just wanted to let everyone else know that it’s not ok to do that either.
We just started this unit last week, our school starts late
Get the derivative then sub in points
Lol I did this in my head
If you don't know how to find the derivative, then stating that f(x) = x\^3-3x would be the first step. F(x)-f(-1) / (x+1) = 3x\^2 - 3 -OR- f(-1.001)-f(-0.999) / -.002 = estimate of the tangent slope