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2. We have used the definition of the derivative at a point. We can also write the derivative of f(x) as a new function called f'(x). For any point a = a the new function f' gives the slope of the line that is tangent to the original function f at a = a. Think of it this way: f(x) and f'(x) are two different pieces of information that you can glean from looking at the graph of y = f(x): f(a) is the height of the point (a, f(a)) f'(a) is the slope of the line tangent to the curve at the point (a, f(x)) Let's see how that works in practice. (a) The following is the graph of a function y = f(x). At the values r = -4,-3, -2,-1,0, 1, 2, 3 and 4, draw a tangent line to the function and visually estimate the slope of the line. Example: it looks like the tangent line at z = 2 has a slope of about 1.0, so I have filled in that value in the table. Use your estimates to complete this table: -4 -3 -1 0 1 2 3 4 1.0 Slope of tangent || -2 -1 (b) The values in the bottom line of the table are the values of f'(x). For example, the table shows that f'(2) = 1.0. Plot these points in the ry-plane, and sketch a line or curve through. them. This is the graph of y = f'(a). L₁_1Y=f(x) (e) Now, try the same exercise with this function g(x). Make a table of slope values (for the integers from -3 up to 4), plot them in the zy-plane, and sketch a line or curve through them to find the graph of y = g(x). Y₁ = g(x) -b - 3

2. We have used the definition of the derivative at a point. We can also write the derivative of f(x) as a new function called f'(x). For any point a = a the new function f' gives the slope of the line that is tangent to the original function f at a = a. Think of it this way: f(x) and f'(x) are two different pieces of information that you can glean from looking at the graph of y = f(x): f(a) is the height of the point (a, f(a)) f'(a) is the slope of the line tangent to the curve at the point (a, f(x)) Let's see how that works in practice. (a) The following is the graph of a function y = f(x). At the values r = -4,-3, -2,-1,0, 1, 2, 3 and 4, draw a tangent line to the function and visually estimate the slope of the line. Example: it looks like the tangent line at z = 2 has a slope of about 1.0, so I have filled in that value in the table. Use your estimates to complete this table: -4 -3 -1 0 1 2 3 4 1.0 Slope of tangent || -2 -1 (b) The values in the bottom line of the table are the values of f'(x). For example, the table shows that f'(2) = 1.0. Plot these points in the ry-plane, and sketch a line or curve through. them. This is the graph of y = f'(a). L₁_1Y=f(x) (e) Now, try the same exercise with this function g(x). Make a table of slope values (for the integers from -3 up to 4), plot them in the zy-plane, and sketch a line or curve through them to find the graph of y = g(x). Y₁ = g(x) -b - 3

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter4: Calculating The Derivative
Section4.2: Derivatives Of Products And Quotients
Problem 35E
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This is the complete question 2. Please help me with this.
2. We have used the definition of the derivative at a point. We can also write the derivative of f(x) as a new function called f'(x). For any point x = a the new function f' gives the slope of the line that is tangent to the original function f at x = a. Think of it this way: f(x) and f'(x) are two different pieces of information that you can glean from looking at the graph of y = f(x): f(a) is the height of the point (a, f(a)) f'(a) is the slope of the line tangent to the curve at the point (a, f(x)) Let's see how that works in practice. (a) The following is the graph of a function y= f(x). At the values r = -4, -3, -2, -1,0, 1,2, 3 and 4, draw a tangent line to the function and visually estimate the slope of the line. Example: it looks like the tangent line at x = 2 has a slope of about 1.0, so I have filled in that value in the table. Use your estimates to complete this table: 2 3 4 I -4 -3 -2 -1 1 Slope of tangent 4|−³|-²|-¹|0|¹|₁ 1.0 -4-3-2 Y 5 4 3 2 -1 2 L. 1y = f(x) 5 I (b) The values in the bottom line of the table are the values of f'(2). For example, the table shows that f'(2) = 1.0. Plot these points in the ry-plane, and sketch a line or curve through them. This is the graph of y = f'(x). -4-3-2-1 1 2 3 (c) Now, try the same exercise with this function g(x). Make a table of slope values (for the integers from -3 up to 4), plot them in the zy-plane, and sketch a line or curve through them to find the graph of y = g'(x). Y y = g(x) 5 4 3 (d) In the drawing on the right, one of these is true: either f(x) is the derivative of g(x), or g(x) is the derivative of f(x). Which is it? Explain. Y 5 4 3 2 1 y = f(x) ➖➖➖➖➖ y = g(x) 1 2 3 4 5 6 T
Transcribed Image Text:2. We have used the definition of the derivative at a point. We can also write the derivative of f(x) as a new function called f'(x). For any point x = a the new function f' gives the slope of the line that is tangent to the original function f at x = a. Think of it this way: f(x) and f'(x) are two different pieces of information that you can glean from looking at the graph of y = f(x): f(a) is the height of the point (a, f(a)) f'(a) is the slope of the line tangent to the curve at the point (a, f(x)) Let's see how that works in practice. (a) The following is the graph of a function y= f(x). At the values r = -4, -3, -2, -1,0, 1,2, 3 and 4, draw a tangent line to the function and visually estimate the slope of the line. Example: it looks like the tangent line at x = 2 has a slope of about 1.0, so I have filled in that value in the table. Use your estimates to complete this table: 2 3 4 I -4 -3 -2 -1 1 Slope of tangent 4|−³|-²|-¹|0|¹|₁ 1.0 -4-3-2 Y 5 4 3 2 -1 2 L. 1y = f(x) 5 I (b) The values in the bottom line of the table are the values of f'(2). For example, the table shows that f'(2) = 1.0. Plot these points in the ry-plane, and sketch a line or curve through them. This is the graph of y = f'(x). -4-3-2-1 1 2 3 (c) Now, try the same exercise with this function g(x). Make a table of slope values (for the integers from -3 up to 4), plot them in the zy-plane, and sketch a line or curve through them to find the graph of y = g'(x). Y y = g(x) 5 4 3 (d) In the drawing on the right, one of these is true: either f(x) is the derivative of g(x), or g(x) is the derivative of f(x). Which is it? Explain. Y 5 4 3 2 1 y = f(x) ➖➖➖➖➖ y = g(x) 1 2 3 4 5 6 T
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