Podcast
Questions and Answers
At what value of x for x > 0 does the line tangent to the graph of f at x have slope 2?
At what value of x for x > 0 does the line tangent to the graph of f at x have slope 2?
2.287
What is the difference between the approximation and the actual value of f′(0.5)?
What is the difference between the approximation and the actual value of f′(0.5)?
0.433
What is the result of f′(0.4) for the function f(x) = (1/7)x^7 + 12x^6 − x^5 − (15/4)x^4 + (4/3)x^3 + 6x^2?
What is the result of f′(0.4) for the function f(x) = (1/7)x^7 + 12x^6 − x^5 − (15/4)x^4 + (4/3)x^3 + 6x^2?
f′(0.4)
Study Notes
Derivative and Tangent Line
- The derivative function is expressed as f′(x) = 0.1x + e^(0.25x).
- For x > 0, a tangent line to the graph of f has a slope of 2.
- Upon solving the equation f′(x) = 2, the value of x is determined to be approximately 2.287.
Function Approximation
- The function f is defined as f(x) = 2x³.
- To approximate f′(0.5) using table values, a difference quotient is applied.
- The calculated derivative at x = 0.5 is f′(0.5) = 0.567.
- An estimate derived from the table yields a derivative approximation of 1, resulting in an error of 0.433 between the actual and estimated values.
Function Analysis
- The function f is represented by f(x) = (1/7)x^7 + 12x^6 − x^5 − (15/4)x^4 + (4/3)x^3 + 6x^2.
- An important value in this context is f′(0.4), indicating the evaluation of the derivative at x = 0.4.
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Description
Test your understanding of derivative concepts and tangent lines with this multiple-choice quiz. This quiz focuses on determining slopes and evaluating functions in the context of calculus. Perfect for students studying Unit 2 of their calculus curriculum.