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Chapter 7: Problem 21

Use the limit definition to find the slope of the tangent line to the graph of \(f\) at the given point. $$ f(x)=x^{3}-x ;(2,6) $$

Short Answer

Expert verified
The slope of the tangent line to the graph of the function \(f(x)=x^{3}-x\) at the point (2,6) is 11.

Step by step solution

01

Identify the function and the point

The function is given by \(f(x)=x^{3}-x\) and the point of interest is \(x=2\).
02

Compute the derivative using the limit definition

According to the limit definition, the derivative of the function \(f(x)\) is computed as follows: \[f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\] So, substitute \(f(x)=x^{3}-x\) and simplify, your expression to compute will be: \[f'(x) = \lim_{h \to 0} \frac{[(x+h)^3-(x+h)] - [x^3-x]}{h}\] After simplifying this expression, we arrive at \(f'(x) = 3x^2-1\).
03

Evaluate the derivative at the given point

Now, it is necessary to evaluate \(f'(x)\) at \(x=2\) to get the slope of the tangent line at this point. Hence, compute \(f'(2)\) by replacing \(x\) with \(2\) in \(f'(x)=3x^2-1\), we get: \(f'(2) =3(2)^2 - 1 = 11\).

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