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Chapter 3: Problem 41

Let \(f(x)=2 x^{3}-3 x^{2}-12 x+4\). a. Find all points on the graph of \(f\) at which the tangent line is horizontal. b. Find all points on the graph of \(f\) at which the tangent line has slope 60.

Short Answer

Expert verified
b) At which points on the graph does the tangent line have a slope of 60? Answer: a) The points on the graph with a horizontal tangent line are \((-1,-9)\) and \((2,4)\). b) The points on the graph with a tangent line slope of 60 are \((-3,61)\) and \((4,44)\).

Step by step solution

01

1. Finding the derivative of f(x)

To find the derivative of \(f(x) = 2x^3 - 3x^2 - 12x + 4\), we will apply the power rule for differentiation: $$f'(x) = \frac{d(2x^3 - 3x^2 - 12x + 4)}{dx} = 6x^2 - 6x - 12$$
02

2. Finding the points with a horizontal tangent line

We set the derivative equal to 0 and solve for x to find the points where the tangent line is horizontal: $$f'(x) = 0$$ $$6x^2 - 6x - 12 = 0$$ Solving this quadratic equation, we find that: $$x = -1, \, 2$$ To find the points on the graph, we substitute these x values back into the original function f(x): $$f(-1) = -9,\, f(2) = 4$$ Therefore, the points on the graph with a horizontal tangent line are \((-1,-9)\) and \((2,4)\).
03

3. Finding the points with tangent line slope 60

We set the derivative equal to 60 and solve for x to find the points with a tangent line slope of 60: $$f'(x) = 60$$ $$6x^2 - 6x - 12 = 60$$ Subtract 60 from both sides: $$6x^2 - 6x - 72 = 0$$ Divide by 6: $$x^2 - x - 12 = 0$$ Solving this quadratic equation, we find that: $$x = -3, 4$$ To find the points on the graph, we substitute these x values back into the original function f(x): $$f(-3) = 61,\, f(4) = 44$$ Therefore, the points on the graph with a tangent line slope of 60 are \((-3,61)\) and \((4,44)\).

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