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Determine the value of the “minimum function” of a function

by RoRi

Let

    \[ f(x) = -\frac{1}{3} x^3 + t^2 x \]

for each t \in \mathbb{R}. Let m(t) denote the minimum of f(x) for x \in [0,1]. Determine m(t) for each t \in [-1,1]. The minimum of the function may occur at the endpoints 0 and 1 in some cases.

Since Apostol specifically warns us to watch out for the minimum solution being on the endpoints of [0,1] we calculate the value at 0 and 1,

    \begin{align*} f(0) &= 0 \\ f(1) &= t^2 - \frac{1}{3}. \end{align*}

Now, we take the derivative

    \[ f'(x) = - x^2 + t^2. \]

Setting this equal to 0, we find that if f(x) has an extremum in the interior of [0,1] then it occurs at

    \[ x^2 = t^2 \quad \implies \quad x = |t| \qquad (\text{since } x \geq 0.) \]

Then, evaluating f(x) at x = |t| we have

    \[ f(|t|) = -\frac{1}{3} t^2 |t| + t^2 |t| = \frac{2}{3} t^2 |t|. \]

But this is greater than or equal to 0 for all values of t \in [-1,1], so it cannot be the minimum (since we know f(0) = 0, so it is at least greater than f(0)). Therefore, the minimum must occur on the end points. From our evaluation of f(0) and f(1) we have

    \[ m(t) = \begin{dcases} 0 & \text{if } t^2 \geq \frac{1}{3} \\ t^2 - \frac{1}{3} & \text{if } t^2 < \frac{1}{3} \end{dcases}. \]

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