Find an equation of the line tangent to f(x)=3x^3-5x at x=2

Find an equation of the line tangent to f(x)=3x^3-5x at x=2.

Thank you very much!

Determine the point by substituting x with 2 for f(x):

f(2) = 3(2)³ - 5(2)
= 3(8) - 10
= 24 - 10
= 14

The point is (2,14). Determine the slope by differentiating f(x) and substituting the same x value.

f'(x) = 9x² - 5
f'(2) = 9(2)² - 5
= 9(4) - 5
= 36 - 5
= 31

Finally, by y - y1 = m(x - x1), letting m = 31 and (x1,y1) = (2, 14):

y - 14 = 31(x - 2)
y - 14 = 31x - 62
y = 31x - 62 + 14
y = 31x - 48

I hope this helps!

Slope of a tangent line is equal to the derivative of the curve at the point of tangency.
f(x) = 3x³ - 5x
f'(x) = 9x² - 5

f(2) = 24-10 = 14
f'(2) = 36-5 = 31

Line through (2, 14) with slope 31

(y-14)/(x-2) = 31
y = 14 + 31(x-2)
y = 14 + 31x - 62

y = 31x - 48