Вопрос школьника по предмету Математика
Дана функция f(x)= e в степени-2 *x в квадрате(если расписать по матем.,то e^(-2*x^2) ),x0=0,найти первую и вторую производную.Вычислить значения этих. производных в заданной точке Хо=0 Найти уравнения касания и нормали в точке Хо=0
Ответ учителя по предмету Математика
f(x)=e^(-2x²)
f(x0)=e^(-2*0²)=1
f'(x)=e^(-2x²)*(-4x)
f'(0)=e^(-2*0²)*(-4*0)=0
f»(x)=e^(-2x²)*(-4x)²-4e^(-2x²)=4*e^(-2x²)(4х²-1)
f»(0)=e^(-2*0²)*(-4*0)²-4e^(-2*0²)=0
общее уравнение касательной в точке х0:
f(x0)=f'(x0)*(x-x0)+f(x0)
f(x0)=0*(x-0)+1=1
f(x) в точке х0 имеет вид
f(x0)=1
уравнение нормали
f(x0)=f(x0)-1/f'(x0)*(x-x0)
f(0)=1-1/0*(x-0) , уравнение составить невозможно, деление на 0.
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