Problem 44 - College Algebra and Calculus: An Applied Approach

Question

Find an equation of the tangent line to the graph of \(f\) at the point \((2, f(2)) .\) Use a graphing utility to check your result by graphing the original function and the tangent line in the same viewing window. $$ f(x)=x \sqrt{x^{2}+5} $$

Step-by-Step Solution

Answer

The equation of the tangent line at the point (2, f(2)) for the function \(f(x) = x \sqrt{x^2 + 5} \) is y = 4x - 2.

Step 1: Calculate the derivative f'(x)

The function f(x) = x \sqrt{x^2 +5}. The derivative can be calculated using the product and chain rules of differentiation.Step 1 (a): Define the two functions for the product rule, \( u = x \) and \( v = \sqrt{x^2+5} \) Step 1 (b): Find the derivatives of u and v. The derivative \( u' = 1 \). To find the derivative of v, which is \( v(x) = (x^2 + 5)^{1/2} \), we can use the chain rule. The derivative \( v' = \frac{1}{2}(x^2 + 5)^{-1/2}(2x) = x(x^2 + 5)^{-1/2} \)Step 1 (c): Find the derivative f'(x) using the product rule formula \( (uv)' = u'v + v'u \). Thus, f'(x) = \( 1*\sqrt{x^2 + 5} + x * x(x^2 + 5)^{-1/2} \)\( = 1*\sqrt{x^2 + 5} + x^2(x^2 + 5)^{-1/2} \)\( = (x^2 + 5)^{1/2} + \frac{x^2}{\sqrt{x^2 + 5}}.\)

Step 2: Find the slope at the given point

The slope of the tangent line at the point (2, f(2)) is the derivative evaluated at x=2. f'(2) = \( (2^2 + 5)^{1/2} + \frac{2^2}{\sqrt{2^2 + 5}} \)\( = \sqrt{9} + \frac{4}{\sqrt{9}} \)\( = 3 + \frac{4}{3} = 4.\)

Step 3: Find the equation of the tangent line

We have the slope of the tangent line \( m = 4 \), a point on the line \( (x_1, y_1) = (2, f(2)) \). Now, we can use the point-slope form of the line to calculate the equation of the tangent line.The point-slope form is given by \( y - y_1 = m(x - x_1) \). Substituting our known values, we get \( y - f(2) = 4(x - 2) \). Therefore the equation of the tangent line is:y = 4x - 8 + f(2), where f(2) = 2*sqrt((2^2) + 5) = 2*sqrt(9) = 6.So, the equation of the line is y = 4x - 2.

Step 4: Verification

As a final step, the original function f(x) = xsqrt(x^2+5) and the derived tangent line equation y=4x-2 can be graphed using a graphing utility to verify the solution. The line should be tangent to the curve at the point (2, f(2)).