9 Calculate the coordinates of the point on the curve $$ y=(2-5 x)^{3}+1 $$ at which the gradient is 0 .

Question

UpStudy AI · Accepted Answer

To find the coordinates of the point on the curve $$ y = (2 - 5x)^3 + 1 $$ where the gradient (derivative) is zero, follow these steps:

1. **Find the derivative of $$ y $$ with respect to $$ x $$:**

$$
   y = (2 - 5x)^3 + 1
   $$
   
   Using the chain rule:

$$
   \frac{dy}{dx} = 3(2 - 5x)^2 \cdot (-5) = -15(2 - 5x)^2
   $$

2. **Set the derivative equal to zero to find the critical points:**

$$
   -15(2 - 5x)^2 = 0
   $$
   
   $$
   (2 - 5x)^2 = 0
   $$
   
   $$
   2 - 5x = 0 \quad \Rightarrow \quad x = \frac{2}{5}
   $$

3. **Find the corresponding $$ y $$-coordinate by substituting $$ x = \frac{2}{5} $$ back into the original equation:**

$$
   y = \left(2 - 5 \cdot \frac{2}{5}\right)^3 + 1 = (2 - 2)^3 + 1 = 0 + 1 = 1
   $$

4. **Thus, the point where the gradient is zero is:**

$$
   \left( \frac{2}{5},\ 1 \right)
   $$

**Final Answer:**  
The point with zero gradient is (2⁄5, 1).
The point where the gradient is zero is (2⁄5, 1).

9 Calculate the coordinates of the point on the curve \( y=(2-5 x)^{3}+1 \) at which the gradient is 0 .