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Determine the point(s) (if any) at which the graph of the function has a horizontal tangent line. $$ y=x^{4}-8 x^{2}+2 $$

Short Answer

Expert verified
The points at which the graph of the function \( x^{4}-8 x^{2}+2 \) has a horizontal tangent line are \( (0,2), (2,-6), (-2,-6) \).

Step by step solution

01

Differentiation

Differentiate the given function with respect to x to get the derivative. Using the power rule, the derivative \( y' = 4 x^{3} - 16 x \).
02

Set Derivative to Zero

Horizontal tangent lines occur where the slope of the tangent, i.e., the derivative is zero. Set the derivative equal to zero and solve for x. \( 4 x^{3} - 16 x = 0 \).
03

Factor the Equation

Factor the equation to simplify it and find the roots. It can be factored by taking out the common factor \(4x\), which gives \( 4x(x^{2} - 4) = 0 \).
04

Solve for x

Find the values of x which satisfy the factorized equation. There are three solutions: \( x = 0, x = 2, x = -2 \).
05

Find corresponding y values

Substitute these x values back into the original function to find the corresponding y values. The points are \( (0,2), (2,-6), (-2,-6) \)

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Most popular questions from this chapter

In Exercises \(81-88\), (a) find an equation of the tangent line to the graph of \(f\) at the indicated point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. \(\frac{\text { Function }}{y=\cos 3 x} \quad \frac{\text { Point }}{\left(\frac{\pi}{4},-\frac{\sqrt{2}}{2}\right)}\)

Linear and Quadratic Approximations The linear and quadratic approximations of a function \(f\) at \(x=a\) are \(P_{1}(x)=f^{\prime}(a)(x-a)+f(a)\) and \(P_{2}(x)=\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2}+f^{\prime}(a)(x-a)+f(a)\) \(\begin{array}{llll}\text { In Exercises } & 133-136, & \text { (a) find the specified linear and }\end{array}\) quadratic approximations of \(f,\) (b) use a graphing utility to graph \(f\) and the approximations, (c) determine whether \(P_{1}\) or \(P_{2}\) is the better approximation, and (d) state how the accuracy changes as you move farther from \(x=a\). $$ \begin{array}{l} f(x)=\sec 2 x \\ a=\frac{\pi}{6} \end{array} $$

In Exercises \(81-88\), (a) find an equation of the tangent line to the graph of \(f\) at the indicated point, (b) use a graphing utility to graph the function and its tangent line at the point, and (c) use the derivative feature of a graphing utility to confirm your results. \(\frac{\text { Function }}{y=2 \tan ^{3} x} \quad \frac{\text { Point }}{\left(\frac{\pi}{4}, 2\right)}\)

Linear and Quadratic Approximations The linear and quadratic approximations of a function \(f\) at \(x=a\) are \(P_{1}(x)=f^{\prime}(a)(x-a)+f(a)\) and \(P_{2}(x)=\frac{1}{2} f^{\prime \prime}(a)(x-a)^{2}+f^{\prime}(a)(x-a)+f(a)\) \(\begin{array}{llll}\text { In Exercises } & 133-136, & \text { (a) find the specified linear and }\end{array}\) quadratic approximations of \(f,\) (b) use a graphing utility to graph \(f\) and the approximations, (c) determine whether \(P_{1}\) or \(P_{2}\) is the better approximation, and (d) state how the accuracy changes as you move farther from \(x=a\). \(f(x)=\tan \frac{\pi x}{4}\) \(a=1\)

In Exercises 35 and 36, find an equation of the tangent line to the graph of the equation at the given point. $$ \arcsin x+\arcsin y=\frac{\pi}{2}, \quad\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right) $$

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