Question

question 3 (1 point)
on which interval(s), if any, is the reciprocal of the quadratic function (f(x)=64x^{2}+101) decreasing?
on the interval where f(x) is decreasing.
more information is required.
none of these answers are correct.
the reciprocal is always decreasing on its defined domain.
the reciprocal is always increasing on its defined domain.
on the interval where f(x) is increasing.

Explanation:

Step1: Let \(y = f(x)=64x^{2}+101\) and \(g(x)=\frac{1}{f(x)}=\frac{1}{64x^{2}+101}\).

The function \(y = 64x^{2}+101\) is a parabola opening upwards (\(a = 64>0\)) with vertex at \((0,101)\). Its derivative \(y^\prime=128x\). It is decreasing on the interval \((-\infty,0)\) and increasing on the interval \((0,\infty)\).

Step2: Use the quotient - rule to find the derivative of \(g(x)\).

The quotient rule states that if \(g(x)=\frac{u(x)}{v(x)}\), then \(g^\prime(x)=\frac{u^\prime(x)v(x)-u(x)v^\prime(x)}{v^{2}(x)}\). Here, \(u(x) = 1\), \(u^\prime(x)=0\), \(v(x)=64x^{2}+101\), and \(v^\prime(x)=128x\). So \(g^\prime(x)=\frac{0\times(64x^{2}+101)-1\times128x}{(64x^{2}+101)^{2}}=-\frac{128x}{(64x^{2}+101)^{2}}\).

Step3: Determine where \(g(x)\) is decreasing.

A function \(g(x)\) is decreasing when \(g^\prime(x)<0\). Since \((64x^{2}+101)^{2}>0\) for all real \(x\), we solve the inequality \(-\frac{128x}{(64x^{2}+101)^{2}}<0\). Multiply both sides by \(- 1\) (and reverse the inequality sign) to get \(\frac{128x}{(64x^{2}+101)^{2}}>0\). The denominator is always positive, so the inequality is equivalent to \(128x>0\), which gives \(x > 0\). This is the interval where \(f(x)\) is increasing.

Answer:

On the interval where f(x) is increasing.