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Compute the value of a function satisfying g'(x2) = x3

by RoRi

Define a function g(x) for all x \in \mathbb{R}_{>0} by

    \[ g(1)= 1 \qquad \text{and} \qquad g'(x^2) = x^3 \qquad \text{for all } x > 0. \]

Compute the value g(4).

First, multiplying both sides of the equality in the definition of g by 2x, we have

    \[ g'(x^2) = x^3 \quad \implies \quad 2x g'(x^2) = 2x^4. \]

We want to do this so that on the left, letting f(x) = x^2 we have

    \[ g(f(x))' = (g(x^2))' = 2x g'(x^2). \]

So then we can integrate both sides of the equality,

    \begin{align*} 2x g'(x^2) = 2x^4 && \implies && \int 2xg'(x^2) \, dx &= \int 2x^4 \, dx \\ && \implies && \int (g(x^2))' \, dx &= \frac{2}{5} x^5 + C \\ && \implies && g(x^2) &= \frac{2}{5} x^5 + C. \end{align*}

Now, we use the given initial condition g(1) = 1 to compute C,

    \[ g(1) = 1 = \frac{2}{5}1^5 + C \quad \implies \quad C = \frac{3}{5}. \]

Therefore,

    \[ g(x^2) = \frac{2}{5} x^5 + \frac{3}{5}. \]

So, computing g(4) = g(2^2) we have

    \[ g(4) = \frac{2}{5}2^5 + \frac{3}{5} = \frac{67}{5}. \]

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