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Determine an equation of the line tangent to \( f(x) \) at \( x = 1 \).

Given the function \[ f(x) = x^7 - x^6 - 45x^5 + 45x^4 + 564x^3 - 564x^2 - 1600x + 1600 \]

Answer :

Final Answer:

The equation of the line tangent to f(x) at x=1 is y = 2400x - 2386.

Explanation:

To find the equation of the tangent line to the function f(x) at x = 1, we need to determine both the slope of the tangent line and the point where it touches the curve.

First, we find the derivative of f(x) to get the slope of the tangent line:

f'(x) = 7x⁶ - 6x⁵ - 225x⁴ + 180x³³ + 1692x² - 1128x - 1600.

Now, plug x = 1 into f'(x) to find the slope at x = 1:

f'(1) = 7(1⁶) - 6(1⁵) - 225(1⁴) + 180(1³) + 1692(1²) - 1128(1) - 1600

f'(1) = 7 - 6 - 225 + 180 + 1692 - 1128 - 1600

f'(1) = 138.

So, the slope of the tangent line is 138. To find the point of tangency, plug x = 1 into the original function:

f(1) = 1⁷ - 1⁶ - 45(1⁵) + 45(1⁴) + 564(1³) - 564(1²) - 1600(1) + 1600

f(1) = 1 - 1 - 45 + 45 + 564 - 564 - 1600 + 1600

f(1) = 0.

The point of tangency is (1, 0). Now, we have the slope (m = 138) and a point (1, 0). We can use the point-slope form to find the equation of the tangent line:

y - 0 = 138(x - 1).

Simplifying:

y = 138x - 138.

Further simplifying:

y = 2400x - 2386.

So, the equation of the tangent line to f(x) at x = 1 is y = 2400x - 2386.

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