Find an equation of the tangent line of the function \( g(x)=4 x+\int_{x}^{\frac{\pi}{4}} \sin (2 t) d t \) at \( x=\frac{\pi}{4} \)
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To find the equation of the tangent line for the function \( g(x) = 4x + \int_{x}^{\frac{\pi}{4}} \sin(2t) dt \) at \( x = \frac{\pi}{4} \), we need to first compute \( g\left(\frac{\pi}{4}\right) \) and \( g'\left(x\right) \). 1. First, we evaluate \( g\left(\frac{\pi}{4}\right) \): \[ g\left(\frac{\pi}{4}\right) = 4\left(\frac{\pi}{4}\right) + \int_{\frac{\pi}{4}}^{\frac{\pi}{4}} \sin(2t) dt = \pi + 0 = \pi. \] 2. Next, we find the derivative \( g'(x) \): By the Fundamental Theorem of Calculus, we have: \[ g'(x) = 4 + \frac{d}{dx} \int_{x}^{\frac{\pi}{4}} \sin(2t) dt = 4 - \sin(2x). \] Now, evaluate \( g'\left(\frac{\pi}{4}\right) \): \[ g'\left(\frac{\pi}{4}\right) = 4 - \sin\left(2 \times \frac{\pi}{4}\right) = 4 - \sin\left(\frac{\pi}{2}\right) = 4 - 1 = 3. \] 3. Now, we have a point \(\left(\frac{\pi}{4}, \pi\right)\) and a slope of 3. The equation of the tangent line in point-slope form is: \[ y - \pi = 3\left(x - \frac{\pi}{4}\right). \] Simplifying this gives: \[ y = 3x - \frac{3\pi}{4} + \pi = 3x + \frac{\pi}{4}. \] Thus, the equation of the tangent line is: \[ y = 3x + \frac{\pi}{4}. \]
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