- Joined
- Jun 27, 2021
- Messages
- 5,386
- Reaction score
- 422
Find the equation of the tangent line to the curve
y = x^3 - x + 5 at the point P (1, 5).
Let me see.
d/dx [x^3 - x + 5] = 2x^2 - 1.
Let x = 1
2(1)^ - 1 = 2 - 1 = 1
The slope is 1.
y = mx + b
Let x = 1, m = 1, and y = 5.
5 = (1)(1) + b
5 = 1 + b
5 - 1 = b
4 = b
y = mx + b
y = (1)x + 4
y = x + 4
Yes?
y = x^3 - x + 5 at the point P (1, 5).
Let me see.
d/dx [x^3 - x + 5] = 2x^2 - 1.
Let x = 1
2(1)^ - 1 = 2 - 1 = 1
The slope is 1.
y = mx + b
Let x = 1, m = 1, and y = 5.
5 = (1)(1) + b
5 = 1 + b
5 - 1 = b
4 = b
y = mx + b
y = (1)x + 4
y = x + 4
Yes?