50.6k views
3 votes
Let h(x) = 5 - 3x^3, h'(3) = ... Use this to find the equation of the tangent line to the curve y=5-3x^3 at the point (3,-76).

User AuRise
by
7.6k points

1 Answer

6 votes

Final answer:

To find the equation of the tangent line to the curve y=5-3x^3 at the point (3,-76), we can use the derivative of the function. The equation of the tangent line is y = -81x + 167.

Step-by-step explanation:

To find the equation of the tangent line to the curve y=5-3x^3 at the point (3,-76), we can use the derivative of the function.

The derivative of h(x) = 5 - 3x^3 is h'(x) = -9x^2.

Now, plug in x = 3 into h'(x) to find the slope of the tangent line: h'(3) = -9(3)^2 = -81.

Since the slope of the tangent line is -81, we can use the point-slope form of a linear equation to find the equation of the tangent line: y - y1 = m(x - x1), where (x1, y1) is the point of tangency.

Plugging in the values: y - (-76) = -81(x - 3)

Simplifying the equation: y = -81x + 243 - 76

Therefore, the equation of the tangent line to the curve y=5-3x^3 at the point (3,-76) is y = -81x + 167.

User Zhafur
by
8.3k points