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Chapter 3: Problem 30

Derivatives and tangent lines a. For the following functions and values of \(a,\) find \(f^{\prime}(a)\) b. Determine an equation of the line tangent to the graph of \(f\) at the point \((a, f(a))\) for the given value of \(a\) $$f(x)=2 x^{3} ; a=10$$

Short Answer

Expert verified
Answer: The equation of the tangent line at the point \((10, 2000)\) is \(y = 600x - 4000\).

Step by step solution

01

Find the derivative of the function

To find the derivative of the function \(f(x) = 2x^3\), we'll use the power rule for derivatives, which states that the derivative of \(x^n\) is \(nx^{n-1}\). Therefore, $$f'(x) = 6x^2$$
02

Find the value of \(f'(a)\)

Now that we have the derivative, we will compute \(f'(a)\) by substituting the given value of \(a\) into the derivative: $$f'(10) = 6(10)^2 = 6(100) = 600$$
03

Calculate the coordinates of the point being considered

In order to find the equation of the tangent line, we need the coordinates of the point \((a, f(a))\). We'll compute this using the given value of \(a\) $$f(a) = f(10) = 2(10)^3 = 2(1000) = 2000$$ So the coordinates of the point are \((10, 2000)\).
04

Determine the equation of the tangent line

Now that we have the coordinates of the point in question and the value of \(f'(a)\), we can find the equation of the tangent line using the point-slope form of a line, which is given by: $$y - y_1 = m(x - x_1)$$ where \((x_1, y_1)\) is the point \((10, 2000)\) and \(m\) is the slope of the tangent line, which is \(f'(a)\). In this case, we have: $$y - 2000 = 600(x - 10)$$ Simplifying the equation of the tangent line, we get: $$y = 600x - 6000 + 2000$$ $$y = 600x - 4000$$ Therefore, the equation of the line tangent to the graph of \(f(x) = 2x^3\) at the point \((10, 2000)\) is: $$y = 600x - 4000$$

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