Find the linearization of f(x) = x^5 at x = 2. Write an exact answer. L(x) = ______

Identify Function and Point: Step 1: Identify the function and the point of linearization. We are given f(x) = x^5 and we need to find its linearization at x = 2. Calculate Derivative: Step 2: Calculate the derivative of f(x). The derivative of f(x) = x^5 is f'(x) = 5x^4. Now, substitute x = 2 into f'(x) to find the slope of the tangent line at that point. f'(2) = 5(2)^4 = 5 * 16 = 80. Calculate Function Value: Step 3: Calculate the function value at x = 2. f(2) = (2)^5 = 32. Write Tangent Line Equation: Step 4: Write the equation of the tangent line, which is the linearization. The formula for the linearization L(x) at a point x = a is L(x) = f(a) + f'(a)(x - a). Substituting the values we found: L(x) = 32 + 80(x - 2).

Home

Math Problems

Calculus

Linear approximation

Full solution

Q. Find the linearization of

f(x) = x^5

x = 2

. Write an exact answer.

\newline

L(x) =

______

Identify Function and Point: Step $1$ : Identify the function and the point of linearization. $\newline$ We are given $f(x) = x^5$ and we need to find its linearization at $x = 2$ .
Calculate Derivative: Step $2$ : Calculate the derivative of $f(x)$ . $\newline$ The derivative of $f(x) = x^5$ is $f'(x) = 5x^4$ . Now, substitute $x = 2$ into $f'(x)$ to find the slope of the tangent line at that point. $\newline$ $f'(2) = 5(2)^4 = 5 \times 16 = 80$ .
Calculate Function Value: Step $3$ : Calculate the function value at $x = 2$ . $f(2) = (2)^5 = 32$ .
Write Tangent Line Equation: Step $4$ : Write the equation of the tangent line, which is the linearization. $\newline$ The formula for the linearization $L(x)$ at a point $x = a$ is $L(x) = f(a) + f'(a)(x - a)$ . $\newline$ Substituting the values we found: $\newline$ $L(x) = 32 + 80(x - 2)$ .