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anonymous
 5 years ago
For the function f, f'(x)= 2x+1 and f(1)=4. What is the approximation for f(1.2) found by using the line tangent to the graph of f at x=1?
anonymous
 5 years ago
For the function f, f'(x)= 2x+1 and f(1)=4. What is the approximation for f(1.2) found by using the line tangent to the graph of f at x=1?

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Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0basically you want to approximate f(1.2) without using integration the idea is that at a point, f' is a farely good approximation of the points around it given that the distance is very close. so the approximation would be more and more off once you go too far. if you wanted to apprx/ f(10) for example, it would be far more off than the apprx. of f(1.2)

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0anyways, you just use a line tangent to f to apprx. in order to do that, you find the linear equation that is tangent to f at that specific point. in your case, the slope of your tangent line can be found by using f'. and it touches f at (1,4) once you find out your linear equation, you plug in x =1.2 are you good so far ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do i find my linear equation?

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0if you know a slope m and a point (a,b) your linear eqn. is yb = m(xa)

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0so you need to find m, and (a,b)

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0m can be found using f' (a,b) is given as the point on f

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so my equation would be y=2x+1?

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.02x+1 = m at any given point. what you should remember is that f is the curve and f' is the "slope that is tangent to f at a given x" so at x =0, f has a tangent line with slope 1 x = 1/2, f has a tangent line with slope 0. since you are approximating the next point around (1,4) you plug this in to see what it is equal to. in your case, x=1 so m =2(1) +1 = 3. sometimes you will have an equation of f' like f' = ycos(x) for those, you have to plug in y as well to find m.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay. i did all that and came up with the answer 4.6..is that correct?

Yuki
 5 years ago
Best ResponseYou've already chosen the best response.0f'(x)= 2x+1 and f(1)=4. What is the approximation for f(1.2) f'(1) = 3 and it goes through (1,4) so the line tangent to f at (1,4) has the linear equation y4 = 3(x1) all you have to do is to plug in 1.2 = x, so y=3(.2)+4 = 4.6 there you go.
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