Two pipes can fill a cistern in 8 and 12 hours respectively. The pipes are opened simultaneously, and it takes 12 minutes more to fill the cistern…

Given $f(x) = x^3 e^{-x}$f(x)=x3e−x, find $f''(1)$f′′(1).

Since $f(x) = x^3 \cdot e^{-x}$f(x)=x3⋅e−x, use the Product Rule with:

• $u = x^3$u=x3 → $u' = 3x^2$u′=3x2
• $v = e^{-x}$v=e−x → $v' = -e^{-x}$v′=−e−x

$$f'(x) = (3x^2)(e^{-x}) + (x^3)(-e^{-x})$$f′(x)=(3x2)(e−x)+(x3)(−e−x)

$$f'(x) = 3x^2e^{-x} - x^3e^{-x}$$f′(x)=3x2e−x−x3e−x

$$f'(x) = e^{-x}(3x^2 - x^3)$$f′(x)=e−x(3x2−x3)

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Differentiate $f'(x) = e^{-x}(3x^2 - x^3)$f′(x)=e−x(3x2−x3) using the Product Rule with:

• $u = e^{-x}$u=e−x → $u' = -e^{-x}$u′=−e−x
• $v = 3x^2 - x^3$v=3x2−x3 → $v' = 6x - 3x^2$v′=6x−3x2

$$f''(x) = (-e^{-x})(3x^2 - x^3) + (e^{-x})(6x - 3x^2)$$f′′(x)=(−e−x)(3x2−x3)+(e−x)(6x−3x2)

$$f''(x) = e^{-x}[-(3x^2 - x^3) + (6x - 3x^2)]$$f′′(x)=e−x[−(3x2−x3)+(6x−3x2)]

$$f''(x) = e^{-x}[-3x^2 + x^3 + 6x - 3x^2]$$f′′(x)=e−x[−3x2+x3+6x−3x2]

$$f''(x) = e^{-x}[x^3 - 6x^2 + 6x]$$f′′(x)=e−x[x3−6x2+6x]

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Evaluate at $x = 1$x=1:

$$f''(1) = e^{-1}[1^3 - 6(1)^2 + 6(1)]$$f′′(1)=e−1[13−6(1)2+6(1)]

$$f''(1) = e^{-1}[1 - 6 + 6]$$f′′(1)=e−1[1−6+6]

$$f''(1) = e^{-1}[1]$$f′′(1)=e−1[1]

$$f''(1) = \dfrac{1}{e}$$f′′(1)=e1​

Therefore, $f''(1) = \dfrac{1}{e}$f′′(1)=e1​