Question

Find the derivative function f′ for the function f. b. Find an equation of the line tangent to the graph of f at (a, f(a)) for the given value of a. 4) f(x)=6x^2−5x−2;a=1

Answer 1

a. The derivative function
`\(f'\)` for the given function
`\(f(x) = 6x^2 - 5x - 2\) is \(f'(x) = 12x - 5\).`

b. The equation of the tangent line to the graph of
`\(f\)` at the point
`\((a, f(a))\) for \(a = 1\) is \(y = 7x - 9\).`

Step-by-step explanation:

a. To find the derivative function
`\(f'\)` of the given function
`\(f(x) = 6x^2 - 5x - 2\),` we apply the power rule and constant rule of differentiation. The derivative of
`\(6x^2\)` is
`\(12x\)` and the derivative of
`\(-5x\) is \(-5\)`, resulting in
`\(f'(x) = 12x - 5\).`

b. To determine the equation of the tangent line at the point
`\((a, f(a))\)` for
`\(a = 1\),` we substitute
`\(x = 1\)` into the derivative function
`\(f'\)`. This gives us the slope of the tangent line at that point. Thus,
`\(f'(1) = 12(1) - 5 = 7\).` Using the point-slope form of a line
`\(y - y_1 = m(x - x_1)\)`, where
`\((x_1, y_1)\)`is the given point and \(m\) is the slope, we obtain the equation of the tangent line as
`\(y = 7x - 9\).`

Understanding the process of finding derivatives and using them to determine the equation of a tangent line is essential in calculus. The derivative represents the rate of change of the function, and the tangent line at a specific point captures the instantaneous rate of change at that point. These concepts are fundamental in analyzing the behavior of functions and their graphs.