Question

21-32 Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative.

\(F\left( t \right) = {t^{\bf{3}}} - {\bf{5}}t + {\bf{1}}\)

Short answer

The derivative is \(F'\left( t \right) = 3{t^2} - 5\). The domain of \(F\left( t \right)\) and \(F'\left( t \right)\) is \(\mathbb{R}\).

Step-by-step solution

Find the derivative of the function by using the definition

The derivative of the function \(F\left( t \right)\) can be calculated as:

\(\begin{aligned}F'\left( t \right) & = \mathop {\lim }\limits_{h \to 0} \frac{{F\left( {t + h} \right) - F\left( t \right)}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {{{\left( {t + h} \right)}^3} - 5\left( {t + h} \right) + 1} \right) - \left( {{t^3} - 5t + 1} \right)}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \frac{{3{t^2}h + 3t{h^2} + {h^3} - 5h}}{h}\\ & = \mathop {\lim }\limits_{h \to 0} \left( {3{t^2} + 3th + {h^2} - 5} \right)\end{aligned}\)

Simplify the equation for \(F'\left( t \right)\)

Simplify the function \(F'\left( t \right) = \mathop {\lim }\limits_{h \to 0} \left( {3{t^2} + 3th + {h^2} - 5} \right)\)to find the derivative:

\(\begin{aligned}F'\left( t \right) & = \mathop {\lim }\limits_{h \to 0} \left( {3{t^2} + 3th + {h^2} - 5} \right)\\ & = 3{t^2} + 0 + 0 - 5\\ & = 3{t^2} - 5\end{aligned}\)

Thus, the derivative is \(F'\left( t \right) = 3{t^2} - 5\).

Find the domain of F and its derivative

As both \(F\left( t \right)\) and \(F'\left( t \right)\) are the polynomials, therefore, the functions are defined for \(t \in \mathbb{R}\).

Thus, the domain of \(F\left( t \right)\) and \(F'\left( t \right)\) is \(\mathbb{R}\).