# Thread: Calculus

1. ## Calculus

Find the coordinates of the point(s) on the graph of:

$

y = \frac{1}{3}x^3 + \frac{1}{2}x^2 - 2x + 1
$

where there are horizontal tangents

Find the x coordinate of the point(s) where:

$

f(x) = 2x\sqrt{2-x^2}
$

has vertical tangents

2. For both questions, you need to calculate the derivative.

The derivative of $\frac{1}{3}x^3 + \frac{1}{2}x^2 - 2x + 1$ is $x^2 + x - 2$, and so the points at which the original function have horizontal tangents are where the derivative is equal to 0. Hence:

$x^2 + x - 2 = 0$

$(x + 2)(x - 1) = 0$

x = -2, 1

To calculate the derivative of $2x\sqrt{2 - x^2}$, use the product rule: $h(x) = f(x)\cdot g(x) \implies h'(x) = f(x)g'(x) + g(x)f'(x)$

Then, to answer the question, determine where the derivative is undefined.

3. thanks!!