Question

Find the maximum and the minimum values of the function f(x) = x²e^x.

Answer 1

f(x) = x²e^x

∴ f'(x) = x²e^x + e^x(2x)

= xe^x(x + 2)

and f"(x) = `d/dx e^x(x^2 + 2x)`

= e^x(2x + 2) + (x² + 2x)e^x

= e^x(2x + 2 + x² + 2x)

= e^x(x² + 4x + 2)

Put f'(x) = 0

∴ xe^x(x + 2) = 0

∴ x(x + 2) = 0, since e^x ≠ 0

∴ x = 0, –2

Now `f^('')(x)|_(x = 0)`

= e⁰(2)

= 2 > 0

∴ f(x) is minimum at x = 0

∴ Min. f(x) = f(0)

= (0)e⁰

= 0

Also, `f^('')(x)|_(x = –2)`

= e^–2(4 – 8 + 2)

= `(-2)/e^2 < 0` 

∴ f(x) is maximum at x = –2

∴ Max. f(x) = f(–2)

= 4e^–2

= `4/e^2`

∴ Minimum value of f(x) = 0

and Maximum value of f(x) = `4/e^2`