# Let f be a function with f(2)=4 and derivative f'(x)=√ (x^3 +1 ) Using a tangent line approximation to graph of f at x=2 estimate f(2.2)?

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- 2 months ago
Find f'(2)

f'(2) = sqrt(2^3 + 1) = sqrt(9) = 3

You need a line with a slope of 3 that passes through (2 , 4)

y - 4 = 3 * (x - 2)

Now find the value of y when x = 2.2

y - 4 = 3 * (2.2 - 2)

y - 4 = 3 * 0.2

y = 4 + 0.6

y = 4.6

That's all there is to it.

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