hint:Albert said:$f(x)$ is a real polynominal function with degree 100,and $f(k)=\dfrac {1}{k} \,\,(k=1,2,3,4,5,------,101),$
please find $f(102)=?$
Albert said:$f(x)$ is a real polynominal function with degree 100,and $f(k)=\dfrac {1}{k} , \,\,(k=1,2,3,4,5,------,101),$
please find $f(102)=?$
kaliprasad said:take $g(x) = x f(x) - 1$
it is polynomial of degree 101 and zero for x = 1 through 101
so $g(x) = A (x-1)(x-2)\cdots(x-101)$
the constant term = $- A * 101! = -1$ so $A = \frac{1}{101!}$
so $g(x) = \frac{1}{101!} (x-1)(x-2)(x-3)\cdots(x-101)$
or $g(102) = 102 f(102) - 1 = -1 $ or $f(102) = 0$
Albert said:please check again $g(102)=?$
The value of $f(102)$ for $f(x)$ of Degree 100 can be found by plugging in 102 for the variable x in the given equation and solving for f(x). This will give you the value of f(102).
Yes, the value of $f(102)$ can be negative for $f(x)$ of Degree 100. The degree of a polynomial does not determine the sign of its value at a specific input. The sign is determined by the coefficients and the value of x.
No, finding the value of $f(102)$ for $f(x)$ of Degree 100 is not the same as finding the x-intercept. The x-intercept is the point where the graph of the function crosses the x-axis, whereas finding the value of $f(102)$ gives the output of the function at a specific input value.
Yes, the value of $f(102)$ for $f(x)$ of Degree 100 can be a fraction. The degree of a polynomial does not determine whether its values are fractions or not. It depends on the coefficients and the value of x.
Finding the value of $f(102)$ for $f(x)$ of Degree 100 is important because it helps us understand the behavior of the function at a specific input value. It can also be used to evaluate the function at other values close to 102 by using methods such as interpolation or extrapolation.