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Multiple Choice
Given the function f(x)=4x2−1, calculate the slope of the tangent line at x=−3.
A
36
B
-36
C
-24
D
24
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Verified step by step guidance
1
To find the slope of the tangent line to the function \( f(x) = 4x^2 - 1 \) at \( x = -3 \), we need to calculate the derivative of the function, \( f'(x) \). The derivative represents the slope of the tangent line at any point \( x \).
Start by differentiating \( f(x) = 4x^2 - 1 \). Use the power rule for differentiation, which states that if \( f(x) = ax^n \), then \( f'(x) = n \, ax^{n-1} \).
Apply the power rule to \( 4x^2 \): \( f'(x) = 2 \, \cdot \, 4x^{2-1} = 8x \). The derivative of a constant, \(-1\), is \(0\). Therefore, \( f'(x) = 8x \).
Now, substitute \( x = -3 \) into the derivative \( f'(x) = 8x \) to find the slope of the tangent line at this point: \( f'(-3) = 8 \, \cdot \, (-3) \).
Calculate \( f'(-3) \) to determine the slope of the tangent line at \( x = -3 \). This will give you the final slope value.