Question

Multiple Choice Assume \(f(x)=\left(x^{2}-1\right)\left(x^{2}+1\right) .\) Which of the following gives the number of horizontal tangents of \(f ?\) \((\mathbf{A}) 0 \quad(\mathbf{B}) 1 \quad(\mathbf{C}) 2 \quad(\mathbf{D}) 3 \quad(\mathbf{E}) 4\)

Short answer

The number of horizontal tangents of the function \(f\) is 1.

Step-by-step solution

Find the derivative of the function

The function is \(f(x) = (x^2 - 1)(x^2 + 1)\). We will use the product rule to find its derivative, \(f'(x)\), where the product rule is \((uv)' = u'v + uv'\). Let's set \(u = x^2 - 1\) and \(v = x^2 + 1\), find their derivatives \(u' = 2x\) and \(v' = 2x\). Substitute into the product rule, we get \(f'(x) = 2x(x^2 + 1) + (x^2 - 1)(2x) = 2x^3 + 2x^3 = 4x^3\)

Solve the equation f'(x) = 0

Now we need to find for which x, the derivative of the function is 0. So we solve the equation \(4x^3 = 0\) for \(x\). Dividing both sides by 4, we get \(x^3 = 0\). Therefore, the solution of the equation is \(x = 0\)

Count the number of solutions

From the solution to the equation, we see that there is only one real number (0) which makes the derivative equal to 0. Therefore, there is only one horizontal tangent to the graph of the function. This corresponds to choice (B): 1.