Let g(x)=xe^(3x). What is the absolute minimum value of g ? Choose 1 answer: (A) -(1)/(3e) (B) (3)/(e^(3)) (C) -(1)/(e^(3)) (D) g has no minimum value

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Let  g(x)=xe^(3x). What is the absolute minimum value of  g ? Choose 1 answer: (A)  -(1)/(3e) (B)  (3)/(e^(3)) (C)  -(1)/(e^(3)) (D)  g has no minimum value

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Find Derivative: To find the absolute minimum value of the function g(x) = xe^(3x), we need to find the critical points by taking the derivative of g(x) and setting it equal to zero. Let's find the derivative g'(x). g'(x) = d/dx [xe^(3x)] Using the product rule, where (u*v)' = u'v + uv', let u = x and v = e^(3x). u' = d/dx [x] = 1 v' = d/dx [e^(3x)] = 3e^(3x) Now apply the product rule. g'(x) = u'v + uv' = 1 * e^(3x) + x * 3e^(3x) = e^(3x) + 3xe^(3x)Set Equal to Zero: Now we set the derivative equal to zero to find the critical points. 0 = e^(3x) + 3xe^(3x) Factor out e^(3x) from both terms. 0 = e^(3x)(1 + 3x) Since e^(3x) is never zero, we can divide both sides by e^(3x) to get: 0 = 1 + 3x Now solve for x. -1 = 3x x = -1/3Critical Point Analysis: We have found a critical point at x = -1/3. To determine if this is a minimum, we can use the second derivative test or analyze the behavior of the first derivative around the critical point. Let's use the second derivative test. Find the second derivative g''(x). g''(x) = d/dx [g'(x)] = d/dx [e^(3x) + 3xe^(3x)] Using the product rule again for the second term, where u = 3x and v = e^(3x), we get: u' = d/dx [3x] = 3 v' = d/dx [e^(3x)] = 3e^(3x) Now apply the product rule. g''(x) = d/dx [e^(3x)] + d/dx [3xe^(3x)] = 3e^(3x) + 3e^(3x) + 3x * 3e^(3x) = 3e^(3x) + 3e^(3x) + 9xe^(3x) = 6e^(3x) + 9xe^(3x)Second Derivative Test: Now we evaluate the second derivative at the critical point x = -1/3. g''(-1/3) = 6e^(-1) + 9(-1/3)e^(-1) = 6/e - 3/e = (6 - 3)/e = 3/e Since 3/e is positive, the second derivative is positive at x = -1/3, which means the function is concave up at this point, and thus x = -1/3 is a local minimum. Because the function g(x) = xe^(3x) goes to negative infinity as x goes to negative infinity and to positive infinity as x goes to positive infinity, this local minimum is also the absolute minimum.Evaluate Second Derivative: Now we need to find the value of g(x) at the critical point x = -1/3 to determine the absolute minimum value. g(-1/3) = (-1/3)e^(3*(-1/3)) = (-1/3)e^(-1) = -(1/3e) This corresponds to answer choice (A).

Let $g(x)=x e^{3 x}$ . $\newline$ What is the absolute minimum value of $g$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $-\frac{1}{3 e}$ $\newline$ (B) $\frac{3}{e^{3}}$ $\newline$ (C) $-\frac{1}{e^{3}}$ $\newline$ (D) $g$ has no minimum value

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Let g(x)=xe3x g(x)=x e^{3 x} g(x)=xe3x.\newlineWhat is the absolute minimum value of g g g ?\newlineChoose 111 answer:\newline(A) −13e -\frac{1}{3 e} −3e1​\newline(B) 3e3 \frac{3}{e^{3}} e33​\newline(C) −1e3 -\frac{1}{e^{3}} −e31​\newline(D) g g g has no minimum value

Let $g(x)=x e^{3 x}$ . $\newline$ What is the absolute minimum value of $g$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $-\frac{1}{3 e}$ $\newline$ (B) $\frac{3}{e^{3}}$ $\newline$ (C) $-\frac{1}{e^{3}}$ $\newline$ (D) $g$ has no minimum value