91影视

To find the extrema of a function, we need to find its derivative. So, let's find the derivatives of f(x) and g(x): For f(x): $$ \begin{aligned} f(x) &= x e^{-x^2} - \int_{0}^{x} e^{-t^2} dt \end{aligned} $$ Applying the Leibniz rule for differentiating under the integral sign and using the chain rule, we get: $$ \begin{aligned} f'(x) &= (1 - 2x^2) e^{-x^2} - e^{-x^2} \end{aligned} $$ For g(x): $$ \begin{aligned} g(x) &= \int_{0}^{x} \frac{\sin t}{1+t^2} dt \end{aligned} $$ Again, applying the Leibniz rule for differentiating under the integral sign, we get: $$ \begin{aligned} g'(x) &= \frac{\sin x}{1+x^2} \end{aligned} $$

We find the critical points by setting the derivatives equal to 0: For f'(x): $$ \begin{aligned} (1 - 2x^2) e^{-x^2} - e^{-x^2} &= 0 \end{aligned} $$ Factor out the common term e^{-x^2}: $$ \begin{aligned} (1 - 2x^2 - 1)e^{-x^2} &= 0 \\ -2x^2 e^{-x^2} &= 0 \end{aligned} $$ Since e^{-x^2} is never 0, we can divide by e^{-x^2} on both sides: $$ \begin{aligned} -2x^2 &= 0 \end{aligned} $$ Solving for x: $$ \begin{aligned} x &= 0 \end{aligned} $$ So, the critical points for f(x) are x=0. For g'(x): $$ \begin{aligned} \frac{\sin x}{1+x^2} &= 0 \end{aligned} $$ The critical points for g(x) occur when sin(x) = 0, which are the multiples of 蟺.

To classify the critical points, we will find the second derivatives and analyze their signs: For f(x), find f''(x): $$ \begin{aligned} f''(x) &= \frac{d}{dx} (-2x^2 e^{-x^2}) \\ &= (-4x + 4x^3) e^{-x^2} \end{aligned} $$ Evaluating f''(0): $$ \begin{aligned} f''(0) &= (-4(0) + 4(0)^3)e^{-0^2} = 0 \end{aligned} $$ Since f''(0) = 0, the second derivative test is inconclusive. However, we can analyze the sign of f'(x) to the left and right of 0: $$ \begin{aligned} f'(x) &= (1 - 2x^2) e^{-x^2} - e^{-x^2} \end{aligned} $$ When x < 0, f'(x) > 0, and when x > 0, f'(x) < 0. This shows that the function f(x) increases to the left of 0 and decreases to the right of 0, so x = 0 is a local maximum for f(x). For g(x), find g''(x): $$ \begin{aligned} g''(x) &= \frac{d}{dx} (\frac{\sin x}{1+x^2}) \\ &= \frac{(1+x^2)\cos x - 2x\sin x}{(1+x^2)^2} \end{aligned} $$ Since g''(x) has a complicated expression, we cannot classify the critical points analytically. However, by observing the graph of g(x), we can determine that the critical points occur at the points where sin(x) = 0 (which are the multiples of 蟺) and change the concavity between those points. Hence, the local maxima and minima of g(x) alternate at the multiples of 蟺, starting with a local maximum at x = 0, a local minimum at x = 蟺, a local maximum at x=-蟺, and continuing in this manner.