Question

If \(f\left( x \right) = 3{x^2} - {x^3}\), find \(f'\left( 1 \right)\) and use it to find an equation of the tangent line to the curve \(y = 3{x^2} - {x^3}\) at the point \(\left( {1,2} \right)\).

Short answer

The equation of the tangent line to the curve \(y = 3{x^2} - {x^3}\) at a point \(\left( {1,2} \right)\) is \(y = 3x - 1\).

Step-by-step solution

The Derivative of a function

The formula of the derivative of a function is shown below:

\(f'\left( a \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {a + h} \right) - f\left( a \right)}}{h}\)

The derivative of a function at \(x = 1\)

Substitute the value \(a = 1\) in the above formula and solve as follows:

\(\begin{aligned}f'\left( 1 \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {1 + h} \right) - f\left( 1 \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{3{{\left( {1 + h} \right)}^2} - {{\left( {1 + h} \right)}^3} - \left( {3{{\left( 1 \right)}^2} - {{\left( 1 \right)}^3}} \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{3 + 6h + 3{h^2} - \left( {1 + {h^3} + 3{h^2} + 3h} \right) - \left( 2 \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} \frac{{3h - {h^3}}}{h}\end{aligned}\)

Solve the above equation further,

\(\begin{aligned}f'\left( 1 \right) &= \mathop {\lim }\limits_{h \to 0} \frac{{h\left( {3 - {h^2}} \right)}}{h}\\ &= \mathop {\lim }\limits_{h \to 0} 3 - {h^2}\\ &= 3 - 0\\ &= 3\end{aligned}\)

The slope of the tangent line to the curve

Recall the fact the derivative \(f'\left( a \right)\) of a function \(y = f\left( x \right)\) is equal to the slope of the tangent line to the curve \(y\) at the point \(x = a\).

Therefore, the slope of the tangent line to the given curve at \(x = 1\) is \(f'\left( 1 \right) = 3\).

The equation of a line

The general equation of a line that passes through the point\(\left( {{x_1},{y_1}} \right)\)and has slope\(m\)is given below:

\(y - {y_1} = m\left( {x - {x_1}} \right)\)

Equation of the tangent line at point \(\left( {1,2} \right)\)

Substitute \(\left( {{x_1},{y_1}} \right) = \left( {1,2} \right)\) and \(m = 3\) in the equation of the line and solve as follows:

\(\begin{aligned}y - {y_1} &= m\left( {x - {x_2}} \right)\\y - 2 &= 3\left( {x - 1} \right)\\y - 2 &= 3x - 3\\y &= 3x - 1\end{aligned}\)

Thus, the equation of the tangent line to the curve \(y = 3{x^2} - {x^3}\) at a point \(\left( {1,2} \right)\) is \(y = 3x - 1\).