Let's work through each part of this question.

Given: The curve $y = x^3 - x^2 + x$ and point $A(1,1)$.

Part (a): Show that $A(1, 1)$ lies on the curve.

To verify that $A(1, 1)$ lies on the curve, substitute $x = 1$ into the equation of the curve and check if $y = 1$.

1. Substitute $x = 1$ into $y = x^3 - x^2 + x$: $y = (1)^3 - (1)^2 + (1) = 1 - 1 + 1 = 1$
2. Since $y = 1$ when $x = 1$, the point $(1, 1)$ lies on the curve.

Part (b): Calculate the gradient of the curve at $A$.

To find the gradient of the curve at $A$, calculate the derivative $\frac{dy}{dx}$ and evaluate it at $x = 1$.

1. Differentiate $y = x^3 - x^2 + x$ with respect to $x$: $\frac{dy}{dx} = 3x^2 - 2x + 1$
2. Substitute $x = 1$ into $\frac{dy}{dx}$ to find the gradient at $A$: $\frac{dy}{dx} \Big|_{x=1} = 3(1)^2 - 2(1) + 1 = 3 - 2 + 1 = 2$

So, the gradient of the curve at $A$ is $2$.

Part (c): Calculate the equation of the tangent to the curve at $A$.

The equation of a tangent line at a point $(x_1, y_1)$ with gradient $m$ is given by: $y - y_1 = m(x - x_1)$

1. Here, $(x_1, y_1) = (1, 1)$ and $m = 2$.
2. Substitute these values into the equation of the tangent: $y - 1 = 2(x - 1)$
3. Simplify the equation: $y = 2x - 2 + 1$ $y = 2x - 1$

So, the equation of the tangent to the curve at $A$ is $y = 2x - 1$.

Would you like more details or have any questions?

Related Questions

1. How do you find the normal to the curve at a given point?
2. What does the second derivative tell us about the curve?
3. How would you find the points where the gradient is zero?
4. How can you determine if the tangent is horizontal or vertical?
5. How would you find the equation of a curve if given its derivative?

Tip:

Always check if a point lies on a curve by substituting its coordinates into the curve's equation.