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For the function
f(x)=-7x^(2)+3x-5, find the slope of the tangent line at
x=3.
Answer:

For the function $f(x)=-7 x^{2}+3 x-5$ , find the slope of the tangent line at $x=3$ . $\newline$ Answer:

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Find the slope of a tangent line using limits

Full solution

Q. For the function

f(x)=-7 x^{2}+3 x-5

, find the slope of the tangent line at

x=3

.

\newline

Answer:

Find Derivative: To find the slope of the tangent line to the function at a specific point, we need to find the derivative of the function, which gives us the slope of the tangent line at any point $x$ . The function is $f(x) = -7x^2 + 3x - 5$ . We will use the power rule for differentiation, which states that the derivative of $x^n$ is $n\cdot x^{(n-1)}$ .
Apply Power Rule: Differentiate the function with respect to $x$ . The derivative of $-7x^2$ is $-14x$ (using the power rule, bringing down the exponent $2$ and subtracting $1$ from the exponent). The derivative of $3x$ is $3$ (since the derivative of $x$ is $1$ ). The derivative of a constant, $-5$ , is $-7x^2$ $0$ (since constants do not change and their derivative is always $-7x^2$ $0$ ). So, the derivative $-7x^2$ $2$ is $-7x^2$ $3$ .
Evaluate at $x=3$ : Now we need to evaluate the derivative at $x = 3$ to find the slope of the tangent line at that point. $\newline$ Substitute $x = 3$ into the derivative $f'(x) = -14x + 3$ . $\newline$ $f'(3) = -14(3) + 3 = -42 + 3 = -39$ .
Calculate Slope: The slope of the tangent line at $x = 3$ is $-39$ .

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