Question 61
What is the derivative of $f(x) = x^61$ at x = 1?
Detailed Solution
f'(x) = 61 $x^{60}$. At x = 1, f'(1) = 61.
✓ Saved
What is the derivative of $f(x) = x^61$ at x = 1?
f'(x) = 61 $x^{60}$. At x = 1, f'(1) = 61.
Consider the function f(x) = (x - $62)^2$. Which of the following statements are TRUE?
f(62) = 0, f'(62) = 0, f''(62) = 2 > 0 (so x = 62 is a local MINIMUM, not maximum).
Find the value of $\lim_{x \to 63} \frac{x^2 - 63^2}{x - 63}$.
Using L'Hopital's rule or factoring: $(x^2 - 63^2)/(x - 63)$ = x + 63. At x = 63, limit = 126.
What is the derivative of $f(x) = x^64$ at x = 1?
f'(x) = 64 $x^{63}$. At x = 1, f'(1) = 64.
Consider the function f(x) = (x - $65)^2$. Which of the following statements are TRUE?
f(65) = 0, f'(65) = 0, f''(65) = 2 > 0 (so x = 65 is a local MINIMUM, not maximum).