Find the linearization L(x) of the function at a = 8. f(x) = x^(2/3)... I got it to be (8/3)x-(4/3)...and its wrong....

f(a)-df(a)(x-a)

....still confuse

f(a) = a^(2/3) = 8^(2/3) = 4 df(x) = (2/3)x^(-1/3) df(a) = 2/3 * 8 ^(-1/3) = 1/3 x-a = x-8 so f(x) = 4-(1/3)(x-8) f(x) = 4 - (x-8)/3 = -x/3 +20/3

all you are doing is finding the tangent line of f(x) at a=8

i had 4/3x-20/3

dhatraditya.....the second part is wrong its df(x)=(2/3)x^(1/3) not -1/3

(8,8^(2/3))=(8,4) is point on the tangent line we can find the slope by finding f'(x) and evaluate f' at x=8 f'(x)=2/3*x^(-1/3) f'(8)=2/3*(8)^(-1/3)=1/3 tangent line has form y=mx+b we know a point and we know the slope so we can find the y-intercept 4=1/3*(8)+b 4-8/3=b 4/3=b so the tangent line is y=1/3*x+4/3

2/3 -1 = -1/3

rmalik, do you understand?

kinda....

L(x)=1/3*x+4/3 is the linearization at a=8 of f(x)=x^(2/3)

all you have to do is find the equation of the tangent line

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