Question

Linearization Find the linearization of \(f(x)=2+\int_{0}^{x} \frac{10}{1+t} d t\) at \(x=0\)

Short answer

The linearization of the function \(f(x) = 2+ \int_{0}^{x} \frac{10}{1+t} d t\) at x = 0 is \(L(x) = 2 + 10x\).

Step-by-step solution

Find the Derivative of the Function

The given function is \(f(x)=2+\int_{0}^{x} \frac{10}{1+t} d t\). The integral part of this function, i.e., \(\int_{0}^{x} \frac{10}{1+t} d t\), can be solved using the Fundamental Theorem of Calculus. This theory states that if a function f is continuous over the interval [a, b] and F is an antiderivative of f on [a, b] then,\(\int_{a}^{b} f(x) dx = F(b) - F(a)\). So, the derivative of \(f(x) = 2+\int_{0}^{x} \frac{10}{1+t} d t\) is \(f'(x) = \frac{10}{1+x}\).

Evaluate the Derivative at the Given Point

We are asked to find the linearization at x = 0. But, for linearization, we need a point (a, f(a)). We already know that a = 0. So, we evaluate \(f'(0) = \frac{10}{1+0}=10\). This gives us m = 10, the slope of the tangent line at x = 0.

Find the Value of f at the Given Point

Next step is to find f(a). At x = 0, the integral becomes 0 because the limits are the same. So, \(f(0) = 2+\int_{0}^{0} \frac{10}{1+t} d t = 2\).

Formulate the Equation of the Tangent Line

The equation of the tangent line at the point a can be expressed as \(L(x) = f(a) + f'(a)(x - a)\). Substituting the values obtained from steps 2 and 3 gives \(L(x) = 2 + 10(x-0)\). Simplifying it, we get the linearization of \(f(x) = 2+\int_{0}^{x} \frac{10}{1+t} d t\) at x = 0 as \(L(x) = 2 + 10x\).