The number of points where the curve y=x^5-20 | vedclass

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(C) Let $f(x) = x^5-20x^3+50x+2$. To find the number of points where the curve crosses the $x$-axis,we analyze the local maxima and minima by finding

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Let $f(x) = x^5-20x^3+50x+2$. To find the number of points where the curve crosses the $x$-axis,we analyze the local maxima and minima by finding the derivative: $f'(x) = 5x^4-60x^2+50 = 5(x^4-12x^2+10)$. Setting $f'(x) = 0$,we get $x^4-12x^2+10 = 0$. Using the quadratic formula for $x^2$: $x^2 = \frac{12 \pm \sqrt{144-40}}{2} = 6 \pm \sqrt{26} \approx 6 \pm 5.1$. So,$x^2 \approx 11.1$ or $x^2 \approx 0.9$. This gives critical points at $x \approx \pm 3.3$ and $x \approx \pm 0.95$. Evaluating the function at these points: $f(-3.3) \approx -100 $f(-0.95) \approx -28 $f(0.95) \approx 32 > 0$ $f(3.3) \approx 104 > 0$ Also,$f(-4)  0$,$f(0) = 2$,$f(2) = -14$,$f(4) > 0$. By the Intermediate Value Theorem,the function changes sign between $(-4, -2)$,$(-2, 0)$,$(0, 2)$,and $(2, 4)$. Thus,there are $5$ points where the curve crosses the $x$-axis.

The number of points where the curve $y=x^5-20x^3+50x+2$ crosses the $x$-axis is $............$.

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