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gulaghasi [49]
3 years ago
10

Please Helppp Cacl AB Determining Slopes.

Mathematics
1 answer:
MrMuchimi3 years ago
4 0
The zeros of the function are those values of x that make y=0. So we solve

0=\dfrac{1+50\sin x}{x^2+3}

The denominator will always be positive, so we can multiply both sides of the equation by it to get

0=1+50\sin x\implies \sin x=-\dfrac1{50}
\implies x=\arcsin\left(-\dfrac1{50}\right)+2n\pi=-\arcsin\dfrac1{50}+2n\pi

where n is any integer. If we take n=\pm1 we should get the two solutions immediately adjacent to the one near x=0 that still lie in the interval -5\le x\le5. So the other two zeros are x=-\arcsin\dfrac1{50}\pm\pi.

The tangent line to the curve at any x is determined by the value of the derivative of the function at that value of x. So first compute the derivative:

y=\dfrac{1+50\sin x}{x^2+3}\implies y'=\dfrac{50\cos x(x^2+3)-2x(1+50\sin x)}{(x^2+3)^2}=\dfrac{(50x^2+150)\cos x-100x\sin x-2x}{(x^2+3)^2}

Now just plug in the values of x determined above. It's helpful to note

\cos\left(\arcsin\dfrac1{50}\right)=\dfrac{7\sqrt{51}}{50}
\cos(\theta+2n\pi)=\cos\theta
\sin(\theta+2n\pi)=\sin\theta
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