Questions: Let h(x)=f(x) g(x). If f(x)=-x^2+x and g(x)=-2 x^2+4 x-1, what is h'(0)?

Transcript text: Let $h(x)=f(x) g(x)$. If $f(x)=-x^{2}+x$ and $g(x)=-2 x^{2}+4 x-1$, what is $h^{\prime}(0) ?$

Solution

Solution Steps

To find the derivative of the function $ h(x) = f(x) g(x) $ at $ x = 0 $, we will use the product rule. The product rule states that if $ h(x) = f(x) g(x) $, then $ h'(x) = f'(x) g(x) + f(x) g'(x) $. We need to:

Compute $ f'(x) $ and $ g'(x) $.
Evaluate $ f(x) $, $ g(x) $, $ f'(x) $, and $ g'(x) $ at $ x = 0 $.
Use the product rule to find $ h'(0) $.

Step 1: Define the Functions

We have the functions: \[ f(x) = -x^2 + x \] \[ g(x) = -2x^2 + 4x - 1 \]

Step 2: Compute the Derivatives

The derivatives of the functions are: \[ f'(x) = 1 - 2x \] \[ g'(x) = 4 - 4x \]

Step 3: Evaluate at $ x = 0 $

Now, we evaluate $ f(0) $, $ g(0) $, $ f'(0) $, and $ g'(0) $: \[ f(0) = -0^2 + 0 = 0 \] \[ g(0) = -2(0)^2 + 4(0) - 1 = -1 \] \[ f'(0) = 1 - 2(0) = 1 \] \[ g'(0) = 4 - 4(0) = 4 \]

Step 4: Apply the Product Rule

Using the product rule: \[ h'(x) = f'(x)g(x) + f(x)g'(x) \] we find $ h'(0) $: \[ h'(0) = f'(0)g(0) + f(0)g'(0) = (1)(-1) + (0)(4) = -1 + 0 = -1 \]