Mathematics 133 – Intensive Calculus for Science 1 Solutions – Discussion 6 November 8, 2000

E. The problem said to find all local maxima and minima of the two functions.

a) f(x) = x^{2}e^{−2x}. Differentiating by the product and chain rules, we have the first derivative

(1)  f′(x) = x^{2}e^{−2x}(−2) + 2xe^{−2x} = (2x − 2x^{2})e^{−2x}

Factoring the polynomial in front, we also can write f′(x) as

(2)  f′(x) = 2x(1 − x)e^{−2x}

Continuing from (1), using the product and chain rules again, the second derivative is:

f′′(x) = (2x − 2x^{2})e^{−2x}(−2) + (2 − 4x)e^{−2x} = (2 − 8x + 4x^{2})e^{−2x}

From the factored form (2), the first derivative is zero at x = 0 and x = 1. Moreover:

f′(x) \{ \begin{matrix}< 0 & if x < 0 \\ > 0 & if 0 < x < 1 \\ < 0 & if x > 1 \end{matrix}

Since f is decreasing to the left of x = 0 and increasing to the right of x = 0 (until x = 1), this says that x = 1 is a local minimum for f. Similarly, since f is increasing to the left of x = 1 and decreasing to the right of x = 1, x = 1 must be a local maximum. (This is the First Derivative Test.) Using f′′(x), we see that f′′(0) = 2 > 0 and f′′(1) = −2e^{−2} < 0. This also shows that x = 0 is a local minimum and x = 1 is a local maximum (Second Derivative Test).

b) f(x) = x^{4} − 6x^{2} + 1. Differentiating we have the first derivative

(3)  f′(x) = 4x^{3} − 12x = 4x(x^{2} − 3)

Continuing from (1), the second derivative is:

f′′(x) = 12x^{2} − 12

From (3), the first derivative is zero at x = 0 and x = ±\sqrt{3}. Moreover:

f′(x) \{ \begin{matrix}< 0 & if x < −\sqrt{3} \\ > 0 & if −\sqrt{3} < x < 0 \\ < 0 & if 0 < x < \sqrt{3} \\ > 0 & if x > \sqrt{3} \end{matrix}

By the First Derivative Test, x = 0 is a local maximum and x = ±\sqrt{3} are local minima. Using f′′(x), we see that f′′(0) = −12 < 0 and f′′(±\sqrt{3}) = 24 > 0. This also shows that x = 0 is a local maximum and x = ±\sqrt{3} are local minima (Second Derivative Test).