What is the equation of the line tangent to $f(x)=2x^3 - x^2-x$ at $x=4$ ?

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Answer 1 · 2017-06-25T00:58:21

$y=15x+48$

This is what we are given:

$f(x)=2x^3 - x^2-x$ at $x=4$

For now we need the y-value which can be calculated simply by solving for $f(4)=108$

$x=4$ $&$ $y=108$

Now we take the derivative using the power rule which states:

$d/dxx^n=nx^(n-1)$

Keep in mind that the derivative of $x$ is just one.

$d/dx=6x-2x-1$

Now we plug in our x-value of $4$ into the derivative:

$6(4)-2(4)-1=15$

This $15$ is our slope of the tangent line.

Now we use the point slope formula which is:

$y-y_1=m(x-x_1)$

$y_1=108$
$m=15$
$x_1=4$

$y-108=15(x-4)$

Solve it:

$y=15x+48$

What is the equation of the line tangent to f(x)=2x^3 - x^2-x f(x)=2x^3 - x^2-x at x=4x=4?