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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basically 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
if you wanted to apprx/ f(10) for example, it would be far more
off than the apprx. of f(1.2)
help me please
anyways, 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 ?
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how do i find my linear equation?
if you know a slope m and a point (a,b)
your linear eqn. is y-b = m(x-a)
so you need to find m, and (a,b)
m can be found using f'
(a,b) is given as the point on f
so my equation would be y=2x+1?
2x+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.
okay. i did all that and came up with the answer 4.6..is that correct?
f'(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
y-4 = 3(x-1)
all you have to do is to plug in 1.2 = x, so
y=3(.2)+4 = 4.6
there you go.