Question

1. - / 1 points details my notes ask your teacher practice another find f(a). f(x) = √(2 - 4x) f(a) = scalcet9 2.xp.7.011. resources read it watch it

Explanation:

Step1: Rewrite the function

Rewrite \( f(x)=\sqrt{2 - 4x} \) as \( f(x)=(2 - 4x)^{\frac{1}{2}} \) using the property of radicals (\( \sqrt[n]{u}=u^{\frac{1}{n}} \)).

Step2: Apply the chain rule

The chain rule states that if \( y = g(h(x)) \), then \( y'=g'(h(x))\cdot h'(x) \). Let \( u = 2-4x \), so \( f(x)=u^{\frac{1}{2}} \). First, find the derivative of \( f(u) \) with respect to \( u \):
Using the power rule \( \frac{d}{du}(u^n)=nu^{n - 1} \), for \( n=\frac{1}{2} \), we have \( f'(u)=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}} \).
Then, find the derivative of \( u = 2-4x \) with respect to \( x \): \( \frac{du}{dx}=- 4 \).
By the chain rule, \( f'(x)=f'(u)\cdot\frac{du}{dx}=\frac{1}{2\sqrt{2 - 4x}}\cdot(-4) \).

Step3: Simplify the derivative

Simplify \( \frac{1}{2\sqrt{2 - 4x}}\cdot(-4) \):
\( \frac{-4}{2\sqrt{2 - 4x}}=\frac{-2}{\sqrt{2 - 4x}} \)

Step4: Evaluate at \( x = a \)

To find \( f'(a) \), substitute \( x = a \) into the derivative:
\( f'(a)=\frac{-2}{\sqrt{2 - 4a}} \)

Answer:

\( \boldsymbol{\frac{-2}{\sqrt{2 - 4a}}} \)