Question

find the equation of the line tangent to the graph of f at (1,4), where f is given by f(x)=2x^3-3x^2+5

Answer 1

`y = 4`

Explanation:

First, find the derivative of f:

`\displaystyle \begin{aligned} f'(x) & = (d)/(dx)\left[ 2x^3 - 3x^2 + 5\right] \\ \\ & = 6x^2 - 6x \\ \\ & = 6x(x-1)\end{aligned}`

Find the instantaneous slope at x = 1:

`\displaystyle \begin{aligned} f'(1) & = 6(1)((1)-1) \\ \\ &= 0 \end{aligned}`

From the point-slope form, find the equation of the line:

`\displaystyle \begin{aligned}y - y_1 & = m(x-x_1) \\ \\ y - (4) & =(0)(x-(1)) \\ \\ y & = 4 \end{aligned}`

In conclusion, the equation of the tangent line is y = 4.