How do you find the equation of the line tangent to #f(x) = x^3# at x = 2?

Question

Ezekiel Alexander · Accepted Answer

#y=2x-16#

#f(2)=2^3=8# and hence #(2,8)# lies on the curve and the tangent.

The gradient (slope) of the tangent at any point is represented by the derivative at that point.

#therefore f'(x)=3x^2#

#therefore f'(2)=3xx2^2=12#.

The tangent is a straight line so has linear equation #y=mx+c#.

Substituting the point #(2,8)# in this equation, we get :

#8=(12)(2)+c#

#therefore c=-16#.

Hence the equation of the required tangent line to the curve is

#y=2x-16#

Charles Anctil · Answer

undefined To find the equation of the line tangent to f(x) = x^3 at x = 2, we need to find the slope of the tangent line and the point of tangency.

First, we find the derivative of f(x) = x^3 using the power rule, which states that the derivative of x^n is n*x^(n-1).

Taking the derivative of f(x) = x^3, we get f'(x) = 3x^2.

Next, we substitute x = 2 into the derivative to find the slope of the tangent line at x = 2.

Substituting x = 2 into f'(x) = 3x^2, we get f'(2) = 3(2)^2 = 12.

So, the slope of the tangent line at x = 2 is 12.

To find the point of tangency, we substitute x = 2 into f(x) = x^3.

Substituting x = 2 into f(x) = x^3, we get f(2) = (2)^3 = 8.

Therefore, the point of tangency is (2, 8).

Using the slope-intercept form of a linear equation, y = mx + b, where m is the slope and b is the y-intercept, we can write the equation of the tangent line.

Substituting the slope (m = 12) and the point of tangency (x = 2, y = 8) into the slope-intercept form, we get y = 12x - 16.

Thus, the equation of the line tangent to f(x) = x^3 at x = 2 is y = 12x - 16.