Questions: Let f(x)=x^2+6x-6 Find f'(−5), f'(0), f'(4). f'(−5) = f'(0) = f'(4) =

Let f(x)=x^2+6x-6 Find f'(−5), f'(0), f'(4). f'(−5) = f'(0) = f'(4) =
Transcript text: Let $f(x)=x^{2}+6 x-6$ Find $f^{\prime}(-5), f^{\prime}(0), f^{\prime}(4)$. \[ \begin{aligned} f^{\prime}(-5) & = \\ f^{\prime}(0) & = \\ f^{\prime}(4) & = \end{aligned} \]
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Solution

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Solution Steps

Step 1: Find the Derivative of \( f(x) \)

The function given is \( f(x) = x^2 + 6x - 6 \). To find the derivative, \( f'(x) \), we apply the power rule to each term:

\[ f'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(6x) - \frac{d}{dx}(6) \]

Using the power rule, \(\frac{d}{dx}(x^n) = nx^{n-1}\), we get:

\[ f'(x) = 2x + 6 \]

Step 2: Calculate \( f^{\prime}(-5) \)

Substitute \( x = -5 \) into the derivative:

\[ f'(-5) = 2(-5) + 6 = -10 + 6 = -4 \]

Step 3: Calculate \( f^{\prime}(0) \)

Substitute \( x = 0 \) into the derivative:

\[ f'(0) = 2(0) + 6 = 0 + 6 = 6 \]

Step 4: Calculate \( f^{\prime}(4) \)

Substitute \( x = 4 \) into the derivative:

\[ f'(4) = 2(4) + 6 = 8 + 6 = 14 \]

Final Answer

\[ \begin{aligned} f^{\prime}(-5) & = \boxed{-4} \\ f^{\prime}(0) & = \boxed{6} \\ f^{\prime}(4) & = \boxed{14} \end{aligned} \]

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