Given the function below Find the equation of the tangent line to the graph of t

Question

Given the function below Find the equation of the tangent line to the graph of the function at x = 1 . Answer in mx + b form. L(x)= Use the tangent line to approximate f(1. 1). L(1. 1)= Compute the actual value of f(1. 1). What is the error between the function value and the linear approximation? Answer as a positive value only. | error | (Approximate to 5 decimal places. ) Box 1: Enter your answer as an expression. Example: 3x^2 + l, x / 5, (a + b) / c Be sure your variables match those in the question Box 2: Enter your answer as a whole or decimal number. Examples: 3, -4, 5. 5 Enter DNE for Does Not Exist, oo for Infinity Box 3: Enter your answer as a whole or decimal number. Examples: 3, -4, 5. 5 Enter DNE for Does Not Exist, oo for Infinity

Explanation / Answer

a) Slope m = dy/dx where y = (-18x^3 + 45)^(1/3)

m = dy/dx = (1/3)[(-18x^3 + 45)^(-2/3)](-18*3*x^2) = -18*(x^2)[(-18x^3 + 45)^(-2/3)]

At x = 1, m = -18*(1^2)[(-18*1^3 + 45)^(-2/3)] = -2

At x = 1, we have y = (-18*1^3 + 45)^(1/3) = 3

Thus equation of line is y-3 = -2(x-1) or L(x) = y = -2x + 5

b) f(1.1) = f(1) + (dy/dx)1*(1.1-1) = 3 + (-2)*(1.1-1) = 2.8

Actual f(1.1) = (-18*1.1^3 + 45)^(1/3) = 2.76076

Error = 2.8 - 2.76076 = 0.03924

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