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- for f(x) = 3-x/2x+3 , find the f'(-1) using the limit definition of derivative. then use it to find an equation of the tangent line to the graph of y=3-x/2x+3 at (-1,4) using f'(a)= lim as h->0 f(a+h)-f(a)/hConsider the function y=g(x)=-x2+3. Use the limit definition of the derivative to compute a formula for y=g′(x).If f(x)=sqrt x, find the linearization L(x) at a=4. Thus the graph of L(x) would be tangent to f(x) at 4.
- for f(x) = 3-x/2x+3 , find the f'(-1) using the limit definition of derivative. then use it to find an equation of the tangent line to the graph of y=3-x/2x+3 at (-1,4)Let f(x) = √x. Find the derivative of f(x) at x = 2 by using the definition of the derivative as a limit.Show that the function f (x, y)=(2 + x-y) / [1+ 2x ^ 2)+(3y ^ 2)] ∈R has a limit at (0,0).
- Let f(x)= 1/(3x+2) find the values below 1) f(x+h)= 2) (f(x+h)-f(x))= 3) lim h→0 (f(x+h)−f(x))/h 4) Find the equation of the line tangent to the graph of f at x=2. y=?For f with derivative as in Figure : (a) Is f (c) a local minimum or maximum? (b) Is f a decreasing function?15) Annual U.S. imports from a certain country in the years 1996 through 2005 could be approximated by I(t) = t2 + 3.5t + 48 (1 ≤ t ≤ 9) billion dollars, where t represents time in years since 1995. Annual U.S. exports to the country in the same years could be approximated by E(t) = 0.5t2 − 1.4t + 13 (0 ≤ t ≤ 10) billion dollars. Assuming that the trends shown in the above models continue indefinitely, calculate the limits lim t→+∞ I(t) and lim t→+∞ I(t)/E(t) algebraically. (If an answer does not exist, enter DNE.) lim t→+∞ I(t) = lim t→+∞ I(t) E(t) = Interpret your answers. In the long term, U.S. imports from the other country will (select) (be rounded or rise without bound) and be times U.S. exports to the other country. Could the given models be extrapolated far into the future? Yes or No
- Let f(x)= √6-x . Compute f'(2) using the limit definition. f'(2)= -1/4 Find an equation of the tangent line at x=2 y= ? How to solve for the equation of the tangent line at x=2?Let f(x) = 3√(x + 1) − 4 = (x + 1)1/3 − 4. Use the limit definition of the derivative to show that f has a vertical tangent line at −1.