Question

In Exercises 67–72, you will explore some functions and their inverses together with their derivatives and tangent line approximations at specified points. Perform the following steps using your CAS:
a. Plot the function y=f(x) together with its derivative over the given interval. Explain why you know that f is one-to-one over the interval.
72. y= 2-x-x³, -2 ≤ x ≤ 2, x_0 = 3/2

Accepted Answer

First, identify the function given: $$y = f(x) = 2 - x - x$$^{3}$$, defined on the interval $$[-2, 2]$$ with the point x_0$ = \frac{3}{2}. $$Calculate the derivative of the function $$f(x)$$ with respect to $$x. $$Use the power rule for derivatives: $$f'(x) = \frac{d}{dx}(2) - \frac{d}{dx}(x) - \frac{d}{dx}(x$$^{3}$$)$$ which simplifies to $$f'(x) = 0 - 1 - 3x$$^{2}$$ = -1 - 3x$$^{2}$$. Plot both the original function f(x$$)$$ and its derivative f$$'(x)$$ over the interval $$[-2, 2]$$. This visual will help you analyze the behavior of the function and its slope at different points. To determine if f$ is one-to-one on the interval, examine the sign of the derivative f$$'(x)$$ over $$[-2, 2]$$. If f$$'(x)$$ does not change sign (always positive or always negative), then f$ is strictly monotonic and therefore one-to-one on that interval. Finally, use the derivative at the point x_0$ = \frac{3}{2} to $$find the slope of the tangent line at that point. Then, write the equation of the tangent line using the point-slope form: $$y - f(x_0$) = f'(x_0$)(x - x_0$).$$

Verified video answer for a similar problem:

Key Concepts

One-to-One Functions

Derivative and Monotonicity

Tangent Line Approximation