Mejia Davison
02/13/2023 · High School
4) Completely Analyze \( K(t)=\left|3 t^{2}-14 t-5\right| \) Reason algebraically, i.e., do not refer to the following graph for your reasoning. Note: \( K(t)=\left\{\begin{array}{cl}3 t^{2}-14 t-5 & 3 t^{2}-14 t-5 \geq 0 \\ -\left(3 t^{2}-14 t-5\right) & 3 t^{2}-14 t-5<0\end{array}\right. \) This will help you get the derivative.
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Step-by-step Solution
To completely analyze the function \( K(t)=\left|3 t^{2}-14 t-5\right| \), we need to consider the two cases separately:
1. When \( 3t^2 - 14t - 5 \geq 0 \):
In this case, the absolute value function is equal to the expression inside the absolute value.
So, \( K(t) = 3t^2 - 14t - 5 \).
2. When \( 3t^2 - 14t - 5 < 0 \):
In this case, the absolute value function is equal to the negative of the expression inside the absolute value.
So, \( K(t) = -(3t^2 - 14t - 5) = -3t^2 + 14t + 5 \).
Now, let's find the derivative of the function \( K(t) \) with respect to \( t \) for each case:
1. For \( K(t) = 3t^2 - 14t - 5 \):
The derivative of \( K(t) \) with respect to \( t \) is:
\[ K'(t) = \frac{d}{dt}(3t^2 - 14t - 5) = 6t - 14 \]
2. For \( K(t) = -3t^2 + 14t + 5 \):
The derivative of \( K(t) \) with respect to \( t \) is:
\[ K'(t) = \frac{d}{dt}(-3t^2 + 14t + 5) = -6t + 14 \]
Therefore, the derivative of the function \( K(t) \) is:
\[ K'(t) = \begin{cases} 6t - 14 & \text{if } 3t^2 - 14t - 5 \geq 0 \\ -6t + 14 & \text{if } 3t^2 - 14t - 5 < 0 \end{cases} \]
Quick Answer
The derivative of \( K(t)=\left|3 t^{2}-14 t-5\right| \) is:
\[ K'(t) = \begin{cases} 6t - 14 & \text{if } 3t^2 - 14t - 5 \geq 0 \\ -6t + 14 & \text{if } 3t^2 - 14t - 5 < 0 \end{cases} \]
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