11. make a rough sketch of the possible shapes of the graph of y=-x^4 + ax^3+bx^2+cx + d where a, b, c, and…

11. make a rough sketch of the possible shapes of the graph of y=-x^4 + ax^3+bx^2+cx + d where a, b, c, and d represent unknown constants. 12. approximate s(2) where s(t)=\\sqrt3{t}-2\\sqrt{t}+\\frac{3}{t}. 13. evaluate the limits in problems 13 - 15. \\lim_{\\delta x\\to0}\\frac{e^{x + \\delta x}-e^{x}}{\\delta x} 14. \\lim_{x\\to2}\\frac{x^{2}+x - 6}{x - 2} 15. \\lim_{x\\to\\infty}\\frac{x}{x + 1} 16. sketch the graph of f(x)=x^{3}+1, and determine the intervals on which f is increasing. 17. solve: 8x-1 = 4 18. find the value of k for which x=-1 is a zero of y = 2x^{3}+x + k. 19. let f(x)=e^{x} and g(x)=-f(-x). graph f and g on the same coordinate plane. 20. let f(x)=\\sin x and g(x)=-3 + 2f(x-\\frac{\\pi}{3}). graph g. 21. find values of x between 0 and 2\\pi such that 2\\sin^{2}x-3\\sin x + 1 = 0. 22. use the definition of the derivative to find f(x) where f(x)=2x^{3}+3x - 4. 23. use a graphing calculator to graph x^{2}+y^{2}-2x + 4y-4 = 0. what are the coordinates of the center of this conic section? 24. find the area of an equilateral triangle whose sides all have length 5. 25. if x = 5 + y, what is the value of x^{2}-2xy + y^{2}?
Answer
Explanation:
Step1: Solve problem 13
Recall the limit - definition of the derivative. The limit $\lim_{\Delta x\rightarrow0}\frac{e^{x + \Delta x}-e^{x}}{\Delta x}$ is the derivative of the function $y = e^{x}$. By the formula for the derivative of the exponential function $(e^{x})'=e^{x}$, so $\lim_{\Delta x\rightarrow0}\frac{e^{x+\Delta x}-e^{x}}{\Delta x}=e^{x}$.
Step2: Solve problem 14
We have the limit $\lim_{x\rightarrow2}\frac{x^{2}+x - 6}{x - 2}$. First, factor the numerator: $x^{2}+x - 6=(x + 3)(x - 2)$. Then $\lim_{x\rightarrow2}\frac{x^{2}+x - 6}{x - 2}=\lim_{x\rightarrow2}\frac{(x + 3)(x - 2)}{x - 2}$. Since $x\neq2$ when taking the limit, we can cancel out the $(x - 2)$ terms, and we get $\lim_{x\rightarrow2}\frac{(x + 3)(x - 2)}{x - 2}=\lim_{x\rightarrow2}(x + 3)=5$.
Step3: Solve problem 15
We have the limit $\lim_{x\rightarrow\infty}\frac{x}{x + 1}$. Divide both the numerator and the denominator by $x$: $\lim_{x\rightarrow\infty}\frac{x}{x + 1}=\lim_{x\rightarrow\infty}\frac{1}{1+\frac{1}{x}}$. As $x\rightarrow\infty$, $\frac{1}{x}\rightarrow0$, so $\lim_{x\rightarrow\infty}\frac{1}{1+\frac{1}{x}} = 1$.
Answer:
- $e^{x}$
- $5$
- $1$