Question

if f(x) = 6x² − x³, find f’(1) and use it to find an equation of the tangent line to the curve y = 6x² − x³ at the point (1, 5).

Explanation:

Step1: Find the derivative of \( f(x) \)

The function is \( f(x) = 6x^2 - x^3 \). Using the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \), we get:
\( f'(x)=\frac{d}{dx}(6x^2)-\frac{d}{dx}(x^3)=6\times2x^{2 - 1}-3x^{3 - 1}=12x - 3x^2 \)

Step2: Evaluate \( f'(1) \)

Substitute \( x = 1 \) into \( f'(x) \):
\( f'(1)=12\times1-3\times1^2=12 - 3 = 9 \)

Step3: Use point - slope form to find the tangent line equation

The point - slope form of a line is \( y - y_1=m(x - x_1) \), where \( (x_1,y_1)=(1,5) \) and \( m = f'(1)=9 \).
Substitute these values into the formula:
\( y - 5=9(x - 1) \)
Simplify the equation:
\( y-5 = 9x-9 \)
\( y=9x - 9 + 5 \)
\( y=9x - 4 \)

Answer:

The equation of the tangent line is \( y = 9x-4 \)