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Find all the values of x where the tangent line is horizontal.
f(x) = 2x^3 + 42x^2 + 270x + 11

1 Answer

4 votes

Answer:

If I calculated correctly, the tangent line is horizontal where x ≈ -5.3 + 9.3i, and -5.3 - 9.3i

I'm somewhat concerned at having gotten complex numbers, and strongly recommend going through the steps to see if I missed anything. I checked it myself and don't see any errors.

Explanation:

You can do this by taking the derivative of the function and solving for zero:

f(x) = 2x³ + 32x² + 220x + 11

f'(x) = 6x² + 64x + 220

f'(x) = 2(3x² + 32x + 110)

We can't factor that further, so let's do it the long way, starting by letting f'(x) equal zero:

0 = 2(3x² + 32x + 110)

0 = 3x² + 32x + 110

0 = 9x² + 96x + 990

0 = 9x² + 96x + 256 + 734

0 = (3x + 16)² + 734

(3x + 16)² = -734

3x + 16 = ± i√734

3x = -16 ± i√734

x = (-16 ± i√734) / 3

x ≈ (-16 + 27.9i) / 3, and (16 - 27.9i) / 3

x ≈ -5.3 + 9.3i, and -5.3 - 9.3i

I'm always wary when I end up with complex numbers. I'd suggest double checking everything here, but I'm fairly certain I did everything correctly.

User Siviwe
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