Question

find the critical numbers of the function. (enter your answers as a comma - separated list. if an answer does not exist, enter dne.)

$h(t)=t^{3 / 4}-3 t^{1 / 4}$

$t=$

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Explanation:

Step1: Differentiate the function

Using the power rule \((x^n)^\prime = nx^{n - 1}\), for \(h(t)=t^{3/4}-3t^{1/4}\), we have \(h^\prime(t)=\frac{3}{4}t^{\frac{3}{4}-1}-3\times\frac{1}{4}t^{\frac{1}{4}-1}=\frac{3}{4}t^{-\frac{1}{4}}-\frac{3}{4}t^{-\frac{3}{4}}\).
Simplify \(h^\prime(t)=\frac{3}{4t^{1/4}}-\frac{3}{4t^{3/4}}=\frac{3t - 3}{4t^{3/4}}\) (by getting a common - denominator \(4t^{3/4}\)).

Step2: Find where \(h^\prime(t) = 0\) or \(h^\prime(t)\) is undefined

• Set \(h^\prime(t)=0\):

Set \(\frac{3t - 3}{4t^{3/4}} = 0\). Since the denominator \(4t^{3/4}
eq0\) for \(t>0\), we solve \(3t - 3 = 0\).
\(3t-3 = 0\) gives \(t = 1\) (add 3 to both sides: \(3t=3\), then divide by 3).

• Find where \(h^\prime(t)\) is undefined:

The function \(h^\prime(t)\) is undefined when \(t = 0\) (because the denominator \(4t^{3/4}=0\) when \(t = 0\)). But we also need to check the domain of the original function \(h(t)\). The domain of \(h(t)=t^{3/4}-3t^{1/4}=\sqrt[4]{t^3}-3\sqrt[4]{t}\) is \(t\geq0\).

Answer:

\(0,1\)