Number of solutions of the equation 3 tan x + | vedclass

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(B) Let $f(x) = 3 	an x + x^3$.First,we find the derivative: $f'(x) = 3 \sec^2 x + 3x^2$.Since $\sec^2 x > 0$ and $x^2 \ge 0$ for all $x$ in the doma

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(B) Let $f(x) = 3 	an x + x^3$.First,we find the derivative: $f'(x) = 3 \sec^2 x + 3x^2$.Since $\sec^2 x > 0$ and $x^2 \ge 0$ for all $x$ in the domain,$f'(x) > 0$ for all $x \in (0, \frac{\pi}{4})$.This implies that $f(x)$ is a strictly increasing function on the interval $(0, \frac{\pi}{4})$.Now,evaluate the function at the endpoints:$f(0) = 3 	an(0) + 0^3 = 0$.$f(\frac{\pi}{4}) = 3 	an(\frac{\pi}{4}) + (\frac{\pi}{4})^3 = 3(1) + \frac{\pi^3}{64} = 3 + \frac{\pi^3}{64}$.Since $\pi \approx 3.14$,$\pi^3 \approx 31$,so $f(\frac{\pi}{4}) \approx 3 + 0.48 = 3.48$.Since $f(0) = 0  2$,and $f(x)$ is continuous and strictly increasing,by the Intermediate Value Theorem,there exists exactly one value $c \in (0, \frac{\pi}{4})$ such that $f(c) = 2$.

Number of solutions of the equation $3 \tan x + x^3 = 2$ in the interval $(0, \frac{\pi}{4})$ is:

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