For the curve f(x) = (x2 + 6x − 5)(1 − x), find the slope of the t

1.	For the curve f(x) = (x2 + 6x − 5)(1 − x), find the slope of the tangent at x = 3.
Answer» Given, f(x) = (x² + 6x − 5)(1 − x) = x² + 6x − 5−x³ − 6x² + 5x = -x³ + 5x² +11x - 5 \(∴ f'(x)=\frac{d}{dx}(-x^3-5x^2+11x+5)\) = −3x² − 5.2x + 11.1 − 0 = −3x² − 10x + 11 ∴ At x = 3, f'(x) = 3.3² − 10.3 + 11 = −27 − 30 + 11 = −57 + 11 = −46 ∴ Slope of the tangent to f(x) at x = 3 is – 46.

For the curve f(x) = (x2 + 6x − 5)(1 − x), find the slope of the tangent at x = 3.

Answer»

Given,

f(x) = (x² + 6x − 5)(1 − x)

= x² + 6x − 5−x³ − 6x² + 5x

= -x³ + 5x² +11x - 5

\(∴ f'(x)=\frac{d}{dx}(-x^3-5x^2+11x+5)\)

= −3x² − 5.2x + 11.1 − 0

= −3x² − 10x + 11

∴ At x = 3,

f'(x) = 3.3² − 10.3 + 11

= −27 − 30 + 11

= −57 + 11

= −46

∴ Slope of the tangent to f(x) at x = 3 is – 46.