The number of real roots of the equation e^{4 | vedclass

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(C) Let $e^{x} = t$. Since $e^{x} > 0$ for all real $x$,we must have $t > 0$.The given equation becomes $f(t) = t^{4} + 2t^{3} - t - 6 = 0$.To find th

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$1$

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(C) Let $e^{x} = t$. Since $e^{x} > 0$ for all real $x$,we must have $t > 0$.The given equation becomes $f(t) = t^{4} + 2t^{3} - t - 6 = 0$.To find the number of real roots,we analyze the function $f(t)$ for $t > 0$.Find the derivative: $f'(t) = 4t^{3} + 6t^{2} - 1$.Find the second derivative: $f''(t) = 12t^{2} + 12t$. For $t > 0$,$f''(t) > 0$,which means $f'(t)$ is strictly increasing for $t > 0$.$f'(0) = -1$ and $f'(1) = 4 + 6 - 1 = 9$. Since $f'(t)$ is continuous and changes sign from negative to positive in the interval $(0, 1)$,there exists a unique root $\alpha \in (0, 1)$ such that $f'(\alpha) = 0$.Thus,$f(t)$ decreases on $(0, \alpha)$ and increases on $(\alpha, \infty)$.Evaluate the function at key points:$f(0) = -6$$f(1) = 1 + 2 - 1 - 6 = -4$$f(2) = 16 + 16 - 2 - 6 = 24$Since $f(1) = -4  0$,by the Intermediate Value Theorem,there is exactly one root for $t$ in the interval $(1, 2)$.Since $t = e^{x} > 0$,and the function $f(t)$ is strictly increasing for $t > 1$,there is exactly one real solution for $x$.

The number of real roots of the equation $e^{4x} + 2e^{3x} - e^{x} - 6 = 0$ is:

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