Use derivatives to find the critical points and inflection points. f(x)=4xe3x

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Use derivatives to find the critical points and inflection points.  f(x)=4xe3x

Nathaniel Kramer · Accepted Answer

1.For any function f, a point x = a in the domain of f where f'(a) = 0 or f'(a) is undefined is called a. critical point of the function. A point, x = c, at which the graph of a continuous function, f, changes concavity is called an inflection point of f.
2.Critical point : Given function $f (x) = 4 x e^{3 x}$ . Taking first derivative, we get.
$f^{'} (x) = \frac{d}{d x} (4 x e^{3 x}) = 12 e^{3 x} x + 4 e^{3 x} = 4 e^{3 x} (3 x + 1)$
Now, $4 e^{3 x} (3 x + 1)$ is defined for all $x \in (- \infty, \infty)$ so critical points are only those points where f(x) =0. Setting (1) to 0, we get
$4 e^{3 x} (3 x + 1) = 0$
This gives us $e^{3 x}$ = 0 or 3x+1=0. However $e^{3 x} \neq q$ 0 so
$3 x + 1 = 0 \Rightarrow x = - \frac{1}{3}$
Therefore $x = \frac{1}{3}$ is the critical point.
3.Inflection point : Taking second derivative from (1), we get
$f^{″} (x) = \frac{d}{d x} (4 e^{3 x} (3 x + 1)) = 36 e^{3 x} x + 24 e^{3 x} = 12 e^{3 x} (824 + 2)$
Now, from definition of inflection point, f(z) must change concavity. The meastire of concavity is given hy f"(c). If "(z) > 0 then f(z) is concave upward at that point and f"(x) <0 then function f(x) is concave downward. In any case, for point of inflection, we must have f#(c) = 0. From (3), we get
$12^{3 x} (3 x + 2) = 0 \Rightarrow x = - \frac{2}{3}$
Now, for x < $- \frac{2}{3}, f x) = 12 e^{3 x} (3 x + 2) < 0$ which is concave downward and for x > 3, f"(x) > 0 so function is concave upward. Therefore f(s) changes concavity at $x = — \frac{2}{3}$ hence it is the inflection point.

Use derivatives to find the critical points and inflection points. f(x)=4xe^{3x}

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