will do. I copied down the answer, but i wrote it down instead so i could grasp what i learned from the screen. now, im trying a similar problem. i will try it myself. however, i got a bit stuck. its the same sort of problem as the one above. f(x) = (x-1) ^2 how do i plug that into f(x+h) ? my guess was [ (x-1) ^2 + h ] not sure though.
(x+h-1)^2 wherever you see x in the given equation is where you will plug in x+h This is for the previous question. just some tips on how to do these (at least the way I do it) Spoiler This might help----- Just remember when you see this type of problem where you have to find the eqn of a tangent line: 1. The first thing you should think of is y-y1=m(x-x1). That is what you are shooting for as your final answer. 2. Fill in the given point into the point slope equation. In this case, (y-(-1)=m(x-0) 3. All you have left to fill in is m, which is the slope, which is the derivative of the function. 4. Find the derivative to the simplest form. That means get rid of all the x's. In this case, the derivative is 4x. You plug in the x from the given point to get a single number without any variables. x=0, 4*0=0 5. plug in the derivative into m in the point slope form. 6. You now have your answer.
thanks for your help rednation, deravusa, new gen, etc. to what you just wrote up there, how exactly does this whole f(x+h) thing work. I struggle with plugging f(x) into f(x+h). I know you just provided the answer to what I was asking about, but the other problems get a bit complicated. for example, skimming through the text book, i see other problems having f(x) as : 1) f(x) = 1/x 2) F(x) = 1/2 x^2 3) f(x) = square root of x+2 when its something simple like f(x) = 3x^2 - 4, its easy to plug that into f(x+h). that would be 3(x+h)^2 - 4, but when the f(x) look more confusing, its hard to plug that into f(x+h)
I'm sorry guys for these stupid questions. I bet they are simple questions for you guys, but this is all new to me. =/
I had the same problem with that when I first did this stuff. All you have to do is find the x in the given equation, and replace it with x+h. 1. 1/(x+h). 2. 1/2(x+h)^2 3. try it yourself and tell us what you got
Get real life tutors or you will fail the class. The questions you ask are really algebra questions - you are not ready for calculus.
YOU HAVE TO LEARN HOW TO LEARN MATH ONCE IT GETS HARD. Knowing a classmate that's learning it with you is super beneficial even if just to understand their approach in learning the new materials. I got through honors geometry and pre-cals and even Calc A in HS fairly easily and barely studied. The stuff just makes sense to me. I think once when trying to figure out an area of a hexagon on a test and didn't remember the formula (remember, don't really study), I was able to just derive the formula based on right triangles and Pythagoras theorem, all in crunch time of needing to finish the test. Am still super proud of myself so many years later Also got a 780/800 on SAT math without ever really prepping for it. But because I never had to work to learn math, never having to pay attention in class, and still do well in them, I actually didn't have the ability to learn math from a classroom setting. So when I actually got to a point in math where it's beyond my abilities I had no idea how to learn it, how to take in and understand new information, I end up really struggling. The thing that got me through it was: 1) Get in study groups with people 2) Go to professor/TA office hours 3) Have them explain in detail you stuff don't really get. Sometimes it's like hitting your head against the wall a million times on one simple thing and suddenly eureka. One last note, it took me Junior year of college to figure out I needed the help (some of that was pride to myself, like it's math, I'm suppose to be good at math). I got through freshman engineering ok, 3.3 gpa or something like that. But Sophomore year was extremely bad and before I knew it, semester was over and my GPA was in shambles (didn't fail any class, but that was a low point). I took a new approach Junior and Senior year and believe when I say it was a good choice.
Why you are talking about plugging in f(x) into f(x+h)? I think you mean "plug in x+h into f". f is a function of a single variable (presumably real-valued). You can therefore "plug in" any real-valued expression through symbolic substitution. For example, if this is f: f(x) = sqrt(x+2) Then f(x+h) just means you symbolically replace x with (x+h): f(x+h) = sqrt((x+h) + 2) = sqrt(x+h+2)
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Goin to professors office hours in a few hours. heres a problem again =/ use the liit definition to find the derivative of the function. Then find the slope of the gangent line to thte graph f at the given point . - f(x) = 4/x ; (1,4) Lim f(x+h) - f(x) / h h-->0 Lim (4 / x+h ) - (4/x) / h h-->0 might be algebra, but not sure how to simplify that further. Afterwards, how do i find the second part of that question , which is to then find the slope of the tangent line to the graph of f at the given point.
^ That problem is no different from the previous one. I suggest you follow the procedure I gave, step by step (including plotting the graph and the tangent line at the given point). Edit: OK, its a bit more tricky, because the resulting formula approaches "0/0" (indeterminate) as h -> 0. First of all, you only care about the value of f'(x) at x=1 (since the point of interest is (1,4)). So, plug in 1 for x. You get: f'(1) ~= [ 4/(1+h) - 4 ] / h, as h approaches 0 Try plugging in increasingly small values for h. h = 0.01: f'(1) ~= [ 4/(1.01) - 4 ] / 0.01 h = 0.001: f'(1) ~= [ 4/(1.001) - 4 ] / 0.001 h = 0.0001: f'(1) ~= [ 4/(1.0001) - 4 ] / 0.0001 ... The idea is that as h gets closer and closer to 0, your computed value for f'(1) becomes more accurate. You'll find that it converges to the value you're looking for (the slope of the tangent line). Once you have that, you know the equation of the line because you already have one point that it crosses through which is (1,4). BTW, if you make the effort to plot the original f(x) and draw the tangent line, that's a very good way to validate the final answer. Not only does it help give you a conceptual feel for what this problem is asking, it will increase the chances of you getting the answer right.
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I prefer doing it numerically rather than graphically. I tried your way, where you graph, but it turns into a mess. I suppose using a graphing calc would be more clear, which i could try. but i want to do the way rednation showed, which was numerically. for your way, i could simply use a graphing calculator, no?
The whole point of these problems if for you to understand the relationship between derivatives and the slope of a tangent line. If your graphing calculator can display the f(x) and a tangent line at a specified point on the curve, then use that as a visual aid as you're working through these problems. But really, you should know how to plot these functions manually yourself (if you don't know how to do it, then take the time to learn). For this case, find the (x,y) points for x = -5, -4, ..., 4, 5. Code: x f(x)=4/x ------------- -5 -4/5 -4 -1 -3 -4/3 -2 -2 -1 -4 0 (infinity) 1 4 2 2 3 4/3 4 1 5 4/5 Then plot those points (x, f(x)) on a x-y graph and connect the dots. There will be two curves, one in the upper-right quadrant and one in the lower-left quadrant. Once you've done that, take a straight edge and draw a line tangent to the upper-right curve at point (1,4). You're looking for the equation of that line. It will be of the form: y = mx + b, and you know that b = 4-m (why?). So, you have the equation of the line being y = mx + 4 - m, and you know that m = f'(1) (why?). I already gave you the procedure of how to calculate f'(1). Substitute that value for m, and you have your equation for the tangent line. If someone knows of a simpler way to answer the problem using the limits definition, please share. That's how I'd do it.
