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#