$2xe ^{(-4x^2)}$
Is this correct?
$f(x) = -4x^2, g(x) = e^x, h(x) = 2x$
$h(x)\cdot g(f(x))$
We are given that
$$f(x) = -4x^2, \quad g(x) = e^x, \quad h(x) = 2x$$
Here is how to express $h(x) \cdot g(f(x))$ that as a combination of elementary functions.
$$\begin{aligned} h(x) \cdot g(f(x)) &= 2x \cdot e^{f(x)}\\ &= 2x \cdot e^{-4x^2} \end{aligned}$$
If we have $h(g(f(x)))$, then
$$\begin{aligned} h(g(f(x))) &= 2g(f(x))\\ &= 2e^{f(x)}\\ &= 2e^{-4x^2} \end{aligned}$$
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We are given that
$$f(x) = -4x^2, \quad g(x) = e^x, \quad h(x) = 2x$$
Here is how to express $h(x) \cdot g(f(x))$ that as a combination of elementary functions.
$$\begin{aligned} h(x) \cdot g(f(x)) &= 2x \cdot e^{f(x)}\\ &= 2x \cdot e^{-4x^2} \end{aligned}$$
If we have $h(g(f(x)))$, then
$$\begin{aligned} h(g(f(x))) &= 2g(f(x))\\ &= 2e^{f(x)}\\ &= 2e^{-4x^2} \end{aligned}$$