Problem 22 - College Algebra and Calculus: An Applied Approach

Question

Use the limit definition to find the slope of the tangent line to the graph of \(f\) at the given point. $$ f(x)=x^{3}+2 x ;(1,3) $$

Step-by-Step Solution

Answer

The slope of the tangent line to the graph of \(f(x) = x^{3} + 2x\) at the point (1,3) is 5.

Step 1: Define the Function

The function is given as \(f(x)=x^{3}+2 x\). This will be used in the limit definition of the derivative.

Step 2: Apply the Limit Definition of the Derivative

Plugging \(f(x)\) into the limit definition, we have \(\lim_{h\to0} \frac{f(x+h) - f(x)}{h}\) or \(\lim_{h\to0} \frac{((x+h)^{3} + 2(x+h)) - (x^{3} + 2x)}{h}\). This simplifies to \(\lim_{h\to0} \frac{3x^2h+3xh^2+h^3+2h}{h}\).

Step 3: Simplify the Expression

Divide the numerator by \(h\) before taking the limit, this gives \(\lim_{h\to0} (3x^2+3xh+h^2+2)\). As \(h\) tends to 0, terms containing \(h\) vanish. Therefore the derivative \(f'(x)\) of the function \(f(x)\) is found to be \(f'(x)=3x^2+2\).

Step 4: Substitute the x-coordinate of the Given Point

To find the slope of the tangent line at the specific point (1,3), compute \(f'(1)\). Substituting \(x=1\) into the derivative gives \(f'(1)=3(1)^2+2=5\). This is the slope of the tangent line at the point (1,3).