Explanation The given function is, f ( x ) = x 2 − 5 x + 5 . Since, the given function is polynomial function. Therefore, the domain of the given function is ( − ∞ , ∞ ) . Substitute 0 for x in the given function to find the value of y − intercept. f ( 0 ) = ( 0 ) 2 − 5 ⋅ 0 + 5 = 0 − 0 + 5 ( ∵ ( 0 ) 2 = 0 ) = 5 Substitute 0 for f ( x ) in the given function to find the value of x − intercept. 0 = x 2 − 5 x + 5 The formula to calculate roots of a quadratic equation is, x = − b ± b 2 − 4 a c 2 a . Substitute − 5 for b , 1 for a and 5 for c in the above equation. x = − ( − 5 ) ± ( − 5 ) 2 − 4 ⋅ 1 ⋅ 5 2 ⋅ 1 = 5 ± 25 − 20 2 = 5 ± 5 2 Solve the values of two roots separately. First root is, x = 5 − 5 2 = 5 − 2.24 2 = 2.76 2 = 1.4 Second root is, x = 5 + 5 2 = 5 + 2.24 2 = 7.24 2 = 3.6 So, x intercepts are ( 1.4 , 0 ) and ( 3.6 , 0 ) and y − intercept is ( 0 , 5 ) . The value of the function f ( x ) at x = ∞ is, lim x → ∞ f ( x ) = lim x → ∞ x 2 − 5 x + 5 = ( ∞ ) 2 − 5 ⋅ ∞ + 5 ( ∵ ( ∞ ) 2 = ∞ ) = ∞ − ∞ + 5 = ∞ The value of the function f ( x ) at x = − ∞ is, lim x → − ∞ f ( x ) = lim x → − ∞ x 2 − 5 x + 5 = ( − ∞ ) 2 − 5 ⋅ ( − ∞ ) + 5 ( ∵ ( − ∞ ) 2 = ∞ ) = ∞ + ∞ + 5 = ∞ Here, f increases without bound as x decreases without bound and that f increases without bound as x increases without bound. Since, the function is polynomial, there are no vertical or horizontal asymptotes. Differentiate the function f and equate to 0 to evaluate the critical point. f ′ ( x ) = 0 d d x x 2 − 5 x + 5 = 0 d d x x 2 − 5 d d x x + d d x 5 = 0 ( ∵ d d x ( x + y + z ) = d d x x + d d x y + d d x z ) 2 x − 5 = 0 ( ∵ d d x x n = n x n − 1 , d d x x = 1 and d d x k = 0 ) Further solve the above equation. 2 x − 5 = 0 2 x = 5 x = 5 2 From above equation, the critical point obtained is x = 5 2

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Author: Tan

Publisher: INTER CENG

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Rate of Change

The relation between two quantities which displays how much greater one quantity is than another is called ratio.

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The change in the vertical distances is known as the rise and the change in the horizontal distances is known as the run. So, the rise divided by run is nothing but a slope value. It is calculated with simple algebraic equations as:

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Chapter 4, Problem 11RE

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Consider the function f(x) = x²-1. (a) Find the instantaneous rate of change of f(x) at x=1 using the definition of the derivative. Show all your steps clearly. (b) Sketch the graph of f(x) around x = 1. Draw the secant line passing through the points on the graph where x 1 and x-> 1+h (for a small positive value of h, illustrate conceptually). Then, draw the tangent line to the graph at x=1. Explain how the slope of the tangent line relates to the value you found in part (a). (c) In a few sentences, explain what the instantaneous rate of change of f(x) at x = 1 represents in the context of the graph of f(x). How does the rate of change of this function vary at different points?

1. The graph of ƒ is given. Use the graph to evaluate each of the following values. If a value does not exist, state that fact. и (a) f'(-5) (b) f'(-3) (c) f'(0) (d) f'(5) 2. Find an equation of the tangent line to the graph of y = g(x) at x = 5 if g(5) = −3 and g'(5) = 4. - 3. If an equation of the tangent line to the graph of y = f(x) at the point where x 2 is y = 4x — 5, find ƒ(2) and f'(2).

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Ch. 4.1 - Explain each of the following: a. f is increasing...Ch. 4.1 - Prob. 2CQ Ch. 4.1 - Prob. 3CQ Ch. 4.1 - Prob. 4CQ Ch. 4.1 - Prob. 5CQ Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E

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