设函数f(x)=g(x)+sinx,曲线y=g(x)在点A( π/ 2 ,g( π /2 ))处的切线方程为y=2x+1,则曲线y=f(x)则曲线y设函数f(x)=g(x)+sinx,曲线y=g(x)在点A( π/ 2 ,g( π /2 ))处的切线方程为y=2x+1,则曲线y=f(x)在点B(π/ 2 ,f(π
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![设函数f(x)=g(x)+sinx,曲线y=g(x)在点A( π/ 2 ,g( π /2 ))处的切线方程为y=2x+1,则曲线y=f(x)则曲线y设函数f(x)=g(x)+sinx,曲线y=g(x)在点A( π/ 2 ,g( π /2 ))处的切线方程为y=2x+1,则曲线y=f(x)在点B(π/ 2 ,f(π](/uploads/image/z/3910285-37-5.jpg?t=%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dg%28x%29%2Bsinx%2C%E6%9B%B2%E7%BA%BFy%3Dg%28x%29%E5%9C%A8%E7%82%B9A%28+%CF%80%2F+2+%2Cg%28+%CF%80+%2F2+%29%29%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B%E4%B8%BAy%3D2x%2B1%2C%E5%88%99%E6%9B%B2%E7%BA%BFy%3Df%EF%BC%88x%EF%BC%89%E5%88%99%E6%9B%B2%E7%BA%BFy%E8%AE%BE%E5%87%BD%E6%95%B0f%28x%29%3Dg%28x%29%2Bsinx%2C%E6%9B%B2%E7%BA%BFy%3Dg%28x%29%E5%9C%A8%E7%82%B9A%28+%CF%80%2F+2+%2Cg%28+%CF%80+%2F2+%29%29%E5%A4%84%E7%9A%84%E5%88%87%E7%BA%BF%E6%96%B9%E7%A8%8B%E4%B8%BAy%3D2x%2B1%2C%E5%88%99%E6%9B%B2%E7%BA%BFy%3Df%EF%BC%88x%EF%BC%89%E5%9C%A8%E7%82%B9B%28%CF%80%2F+2+%2Cf%28%CF%80)
设函数f(x)=g(x)+sinx,曲线y=g(x)在点A( π/ 2 ,g( π /2 ))处的切线方程为y=2x+1,则曲线y=f(x)则曲线y设函数f(x)=g(x)+sinx,曲线y=g(x)在点A( π/ 2 ,g( π /2 ))处的切线方程为y=2x+1,则曲线y=f(x)在点B(π/ 2 ,f(π
设函数f(x)=g(x)+sinx,曲线y=g(x)在点A( π/ 2 ,g( π /2 ))处的切线方程为y=2x+1,则曲线y=f(x)则曲线y
设函数f(x)=g(x)+sinx,曲线y=g(x)在点A( π/ 2 ,g( π /2 ))处的切线方程为y=2x+1,则曲线y=f(x)在点B(π/ 2 ,f(π /2 ))处切线的方程为--------
设函数f(x)=g(x)+sinx,曲线y=g(x)在点A( π/ 2 ,g( π /2 ))处的切线方程为y=2x+1,则曲线y=f(x)则曲线y设函数f(x)=g(x)+sinx,曲线y=g(x)在点A( π/ 2 ,g( π /2 ))处的切线方程为y=2x+1,则曲线y=f(x)在点B(π/ 2 ,f(π
g(x)在A点斜率是2,所以g‘((π /2)=2
f(x)=g(x)+sinx,所以f’(x)=g‘(x)+cosx
f’(π /2)=g‘(π /2)+cos(π /2)=2+0=2,所以f(x)在π /2处导数是2,也就B点切线是斜率是2
g( π /2 )=2(π/ 2)+1=π+1
f(π /2)=g(π /2)+sin(π /2)=π+1+1=π+2,所以B点坐标(π/ 2,π+2)
用点斜式π+ 2+1=2(π/ 2)+b,求得b=2
所以f(x)在B(π/ 2 ,f(π /2 ))切线方程是y=2x+2