Question

find the value of $(f\circ g)$ at the given value.
$f(u)=u^{7}-3$, $u = g(x)=\sqrt{x}$, $x = 1$
$(f\circ g)(1)=\square$
(type an integer or a simplified fraction.)

Explanation:

Step1: Apply chain rule

The chain rule states that \((f\circ g)'(x)=f'(g(x))\cdot g'(x)\).
First, find \(f'(u)\) and \(g'(x)\).
For \(f(u) = u^{7}-3\), using the power rule \((x^{n})'=nx^{n - 1}\), we have \(f'(u)=7u^{6}\).
For \(g(x)=\sqrt{x}=x^{\frac{1}{2}}\), using the power rule, \(g'(x)=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}\).

Step2: Substitute \(u = g(x)\) into \(f'(u)\)

\(f'(g(x))=7(g(x))^{6}=7(\sqrt{x})^{6}=7x^{3}\).

Step3: Calculate \((f\circ g)'(x)\)

\((f\circ g)'(x)=f'(g(x))\cdot g'(x)=7x^{3}\cdot\frac{1}{2\sqrt{x}}\).
Simplify \(7x^{3}\cdot\frac{1}{2\sqrt{x}}=\frac{7}{2}x^{3-\frac{1}{2}}=\frac{7}{2}x^{\frac{5}{2}}\).

Step4: Evaluate at \(x = 1\)

Substitute \(x = 1\) into \((f\circ g)'(x)\): \((f\circ g)'(1)=\frac{7}{2}(1)^{\frac{5}{2}}\).

Answer:

\(\frac{7}{2}\)