Question

1. if f is the function defined by $f(x)=\frac{x^{7}-5 x}{9}$, then the instantaneous rate of change of f at x = 1 is (a) $\frac{2}{9}$ (b) $\frac{4}{9}$ (c) $\frac{7}{9}$ (d) $\frac{10}{9}$

Explanation:

Step1: Recall the formula for the derivative

The instantaneous rate of change of a function \(y = f(x)\) at a point \(x=a\) is given by \(f^{\prime}(a)\). For \(y=\frac{u}{v}\) (where \(u = x^{7}-5x\) and \(v = 9\)), by the quotient rule \((\frac{u}{v})^{\prime}=\frac{u^{\prime}v - uv^{\prime}}{v^{2}}\). Since \(v = 9\), \(v^{\prime}=0\). So \(f^{\prime}(x)=\frac{(7x^{6}-5)\times9-(x^{7}-5x)\times0}{9^{2}}=\frac{7x^{6}-5}{9}\) (using the power rule \((x^{n})^{\prime}=nx^{n - 1}\), so \(u^{\prime}=(x^{7}-5x)^{\prime}=7x^{6}-5\)).

Step2: Evaluate the derivative at \(x = 1\)

Substitute \(x = 1\) into \(f^{\prime}(x)\). We get \(f^{\prime}(1)=\frac{7\times(1)^{6}-5}{9}\).

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Answer:

A. \(\frac{2}{9}\)