Question

Find the function $f$ given that the slope of the tangent line to the graph of $f$ at any point $(x, f(x))$ is $f^{\prime}(x)$ and that the graph of $f$ passes through the given point. $$f^{\prime}(t)=t^{2}-2 t+3 ;(1,2)$$

Name: Problem 60, Chapter 11: Applied Mathematics: For the Managerial, Life, and Social Sciences
Uploaded: 2021-05-10T15:08:37-08:00
Duration: 3 min 9 s
Channel: Gregory Higby
Description: Find the function f given that the slope of the tangent line to the graph of f at any point (x, f(x)) is f^'(x) and that the graph of f passes through the given point. f^'(t)=t^2-2 t+3 ;(1,2)

   Find the function $f$ given that the slope of the tangent line to the graph of $f$ at any point $(x, f(x))$ is $f^{\prime}(x)$ and that the graph of $f$ passes through the given point.
$$f^{\prime}(t)=t^{2}-2 t+3 ;(1,2)$$

Applied Mathematics: For the Managerial, Life, and Social Sciences

Soo T. Tan 5th Edition

Chapter 11, Problem 60

Instant Answer

Solved by Expert Gregory Higby

05/10/2021

Step 1

$$f(t) = \int (t^2 - 2t + 3) dt = \frac{1}{3}t^3 - t^2 + 3t + C$$ Now, we need to find the constant $C$ using the given point $(1, 2)$. We know that $f(1) = 2$, so we can plug in $x = 1$ into the equation: $$2 = \frac{1}{3}(1)^3 - (1)^2 + 3(1) + C$$ Show more…

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Find the function $f$ given that the slope of the tangent line to the graph of $f$ at any point $(x, f(x))$ is $f^{\prime}(x)$ and that the graph of $f$ passes through the given point. $$f^{\prime}(t)=t^{2}-2 t+3 ;(1,2)$$

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Key Concepts

Antiderivative

An antiderivative of a function is another function whose derivative is the original function. In the context of the given problem, finding the antiderivative involves integrating the given derivative to recover the original function up to an unknown constant.

Constant of Integration

When finding an antiderivative, the integration process introduces an arbitrary constant because the derivative of a constant is zero. This constant is determined using additional information provided in the problem, ensuring that the resulting function fits specific conditions.

Initial Condition

An initial condition is a specific value of the function at a given point, which is used to determine the constant of integration. By substituting this value into the antiderivative, the exact function that satisfies both the differential equation and the initial condition can be found.

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