## Equation of a Tangent Line: Problems and Solutions

### 4 What are your thoughts and questions?

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Anonymous

Y=x5 at the point 1,5 give the solution

Matheno

(1) the given equation is y = 5x, in which case the function is a line and so is tangent to itself,

or

(2) the given equation is y = x^5, in which case the point (1,5) isn’t on the curve, but the point (1,1) is.

If you clarify which choice, if either, is the correct question, we’ll try to provide more detail about the solution.

Anonymous

Y=x³at the point 1,1 give the solution

Matheno

Following the approach outlined in the SUMMARY box above:

1. The slope of the tangent line at (1,1) equals the derivative at that point. Since $y=f(x)= x^3,$ the derivative is $y’=f'(x) =3x^2.$ Hence the slope of the tangent line at $x_0 = 1$ is $m_{tangent} = f'(1) = 3.$

2. Then we can write the equation of this tangent line, using the point-slope form of the line that has slope $m_{tangent} = 3$ and contains the point (1,1):
$$y – 1 = 3(x-1)$$

Does that provide the solution you were after? We hope so!