Find the equation of the tangent line to f(x) = x³ + x² + x + 1 at x = 4.

Solution:

The tangent to the curve f(x) = x³ + x² + x + 1 is given by the first differential of f(x).

The differential of the given function is:

f(x) = x³ + x² + x + 1 --- (1)

df(x)/dx = d(x³ + x² + x + 1)/dx

= d(x³)/dx + d(x²)/dx + d(x)/dx + d(1)/dx

= 3x² + 2x + 1 + 0

= 3x² + 2x + 1 --- (2)

So slope of the tangent line at x = 4 is found by substituting x = 4 in (2)

Slope = m = 3(4)² + 2(4) + 1

= 3(16) + 8 + 1

= 48 + 9

m = 57

The value of y coordinate at x = 4 is found by substituting x = 4 in equation (1)

y = (4)³ + (4)² + (4) + 1

= 64 + 16 + 1 = 81

Therefore the equation of the tangent to the curve at the point (4, 81) and slope m = 57 is

(y - 81) = (57)(x - 4)

y - 81 = 57x - 228

y = 57x - 147

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Find the equation of the tangent line to f(x) = x³ + x² + x + 1 at x = 4.

Summary:

The equation of the tangent line to f(x) = x³ + x² + x + 1 at x = 4 is y = 57x - 147