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For the function
f(x)=6x^(2)+11 x+7, find the slope of the tangent line at
x=8.
Answer:

For the function $f(x)=6 x^{2}+11 x+7$ , find the slope of the tangent line at $x=8$ . $\newline$ Answer:

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Find the slope of a tangent line using limits

Full solution

Q. For the function

f(x)=6 x^{2}+11 x+7

, find the slope of the tangent line at

x=8

.

\newline

Answer:

Calculate Derivative: To find the slope of the tangent line to the function at a specific point, we need to calculate the derivative of the function, which gives us the slope of the tangent line at any point $x$ .
Derivative Calculation: The derivative of $f(x) = 6x^2 + 11x + 7$ with respect to $x$ is $f'(x) = \frac{d}{dx} (6x^2) + \frac{d}{dx} (11x) + \frac{d}{dx} (7)$ .
Simplify Derivative: Calculating the derivatives term by term, we get $f'(x) = 2 \times 6x^{2-1} + 11 \times 1 + 0$ , since the derivative of a constant is $0$ .
Evaluate at $x = 8$ : Simplifying the derivative, we get $f'(x) = 12x + 11$ .
Calculate Tangent Slope: Now we need to evaluate the derivative at $x = 8$ to find the slope of the tangent line at that point. So we calculate $f'(8) = 12 \times 8 + 11$ .
Calculate Tangent Slope: Now we need to evaluate the derivative at $x = 8$ to find the slope of the tangent line at that point. So we calculate $f'(8) = 12 \times 8 + 11$ .Performing the calculation, we get $f'(8) = 96 + 11$ .
Calculate Tangent Slope: Now we need to evaluate the derivative at $x = 8$ to find the slope of the tangent line at that point. So we calculate $f'(8) = 12 \times 8 + 11$ .Performing the calculation, we get $f'(8) = 96 + 11$ .Adding the numbers together, we find that $f'(8) = 107$ . This is the slope of the tangent line at $x = 8$ .

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