(-5, 2) (5, 2) Graph of f' The graph of f', the derivative of a function f, consists of two line segments and a semicircle, as shown in the figure above. If f (2) = 1, then f (-5)
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- Find the derivative of: f(z) = (3z+9) / (2-z) 15z + 3z/(2-z)^2 15z – 3z/(2-z)^2 15z/(2-z)^2 15/(2-z)^2 None of the aboveAt the end of the exam, all of a sudden your calculator breaks down. All you have left to do is to compute the value at (2.95,12.2) of the function f(x,y) = √xy, rounded to two decimals. You decide to put to work what you have learned in the math course: • You note that f (3, 12) = 6 and that (2.95, 12.2) is close to (3, 12).• You use the linear approximation of the function f at the point (3, 12) to compute an approxi- mate value for f (2.95, 12.2).• Is your rounded outcome on the exam correct?Give an example of a function f (x) that has one positive derivatives on the range (−1, 0) and a negative derivatives on the range (0, 1).
- Consider the function f(x)=2x^3−12x^2−30x+5 on the interval [−5,7]. Find the slope of the secant line (for this function) whose endpoints are (−5,f(−5)) and (7,f(7)). By the MVT, we know there exists a c in the open interval (−5,7) such that f′(c) is equal to slope of this secant line. Find the two values of c that work.A. Note on Jacobian dterminants and also give economic interpretation of total differentiation. Also discuss the importance of derivatives. Find the derivatives of the following: A) Y = (2x2 + 16) (6x + 10) B) Y = (14x2 + 12) ( 3x-1 - 9)Find a function that isn't a polynomial function or a rational function, that has a non-horizontal point of inflexion at (8,22). That is, you must find a function that has non-zero gradient at that point but second derivative is zero there and the second derivative changes sign at the point. Remembering that your function can't be a polynomial or a rational function.
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- A function f has the tangent line y=2x+1 at the point (3,7). Find the equation of tangent line of f^-1 at the point (7,3) in slope-intercept form.The function f(x) is continuous and differentiable on the interval [2,10]. It is known that f(2)=8 and the derivative on the given interval satisfies the condition f′(x)≤4 for all x∈(2,10). Determine the maximum possible value of the function at x=10.Equation of the line tangent to the graph of f(x)=(x)(1-2x)^3 at (-1,1).