55. Find an equation of the tangent line to the graph of
f(x) =x^3 - 2x + 1
a) at (2,5);
b) at (-1,2);
c) at (0,1).
I believe the answers are:
a) y= (3/2 x) - (3/2)
b) No solution
c) No solution
For each function, find the points on the graph at which the tangent line is horizontal. If none exists, state that fact.
67. y=4
73. f(x) = (1/3)^x^3 + (1/2)^x^2 - 2
For the function, find the points on the graph at which the tangent line has slope 1.
79. y=(1/3)^x^3 + (2)^x^2 + 2x
55. The slope any point of the line is the derivative,
df/dx = 3x^2 -2
When x = 2, f(x) = 5 and
df/dx = (3*4) - 2 = 10
The coordinates of (b) and (c) are also on the line, and there IS a slope there. Use the same df/dx formuyla to compute it.
I will be happy to provide assistance with you other three problems when work is shown. In #79, are you sure that x^3 is a power of 2 and x^2 is a power of 1/2? It looks to me like you used too many ^ signs.
67. dy/dx = 0
Would it be all the points of the function.
To find the equation of the tangent line to a graph at a given point, we can use the derivative of the function.
a) To find the equation of the tangent line to the graph of f(x) = x^3 - 2x + 1 at (2,5), we first need to find the derivative of f(x). Taking the derivative of f(x), we get f'(x) = 3x^2 - 2.
Now, substitute x = 2 into f'(x) to find the slope of the tangent line at the point (2,5). Thus, the slope of the tangent line at (2,5) is f'(2) = 3(2)^2 - 2 = 12.
Using the point-slope form of a linear equation, y - y1 = m(x - x1), where m is the slope and (x1, y1) is the given point, we can substitute the values to find the equation of the tangent line:
y - 5 = 12(x - 2).
Simplifying, we get y - 5 = 12x - 24, which can be rewritten as y = 12x - 19. Therefore, the equation of the tangent line at (2,5) is y = 12x - 19.
b) To find the equation of the tangent line to the graph of f(x) = x^3 - 2x + 1 at (-1,2), we repeat the same process. First, find the derivative of f(x): f'(x) = 3x^2 - 2.
Substitute x = -1 into f'(x) to find the slope of the tangent line at the point (-1,2). The slope is f'(-1) = 3(-1)^2 - 2 = 1.
Using the point-slope form, we have y - 2 = 1(x - (-1)), which simplifies to y - 2 = x + 1. Therefore, the equation of the tangent line at (-1,2) is y = x + 3.
c) To find the equation of the tangent line to the graph of f(x) = x^3 - 2x + 1 at (0,1), we follow the same steps. The derivative of f(x) is f'(x) = 3x^2 - 2.
Substituting x = 0 into f'(x) gives us the slope of the tangent line at the point (0,1). The slope is f'(0) = 3(0)^2 - 2 = -2.
Using the point-slope form, we have y - 1 = -2(x - 0), which simplifies to y - 1 = -2x. Therefore, the equation of the tangent line at (0,1) is y = -2x + 1.
For the next set of questions, we need to find points on the graph where the tangent line is horizontal or has a specific slope.
67. For the equation y = 4, the graph is a horizontal line at y = 4. Since the line is horizontal, there are no points on the graph where the tangent line is horizontal.
73. To find points on the graph of f(x) = (1/3)^x^3 + (1/2)^x^2 - 2 where the tangent line is horizontal, we need to find the derivative of f(x). The derivative of f(x) is f'(x) = (1/3)(1/2)x^2 - 2x + (1/2)(1/2)x^1.
Set f'(x) = 0 and solve for x to find the critical points where the tangent line is horizontal. However, the equation f'(x) = 0 leads to a complex equation, making it difficult to find exact solutions. Therefore, an analytical solution is not possible in this case.
79. To find points on the graph of y = (1/3)^x^3 + 2^x^2 + 2x where the tangent line has a slope of 1, we again need to find the derivative of y. The derivative of y is y' = (1/3)(1/3)x^2 + (2)(2)x + 2.
Set y' = 1 and solve for x to find the critical points where the tangent line has a slope of 1. The equation (1/3)(1/3)x^2 + 4x + 2 = 1 simplifies to (1/9)x^2 + 4x + 1 = 0.
Solve this quadratic equation to find the values of x. Use the quadratic formula, x = (-b ± √(b^2 - 4ac)) / (2a), where a = 1/9, b = 4, and c = 1.
After solving the equation, you should obtain two values for x. Substitute these values back into the original equation y = (1/3)^x^3 + 2^x^2 + 2x to find the corresponding y-values.
Therefore, to summarize the answers:
a) The equation of the tangent line at (2,5) is y = 12x - 19.
b) The equation of the tangent line at (-1,2) is y = x + 3.
c) The equation of the tangent line at (0,1) is y = -2x + 1.
67) There are no points on the graph where the tangent line is horizontal.
73) There is no analytical solution to find the points on the graph where the tangent line is horizontal.
79) Solve the quadratic equation (1/9)x^2 + 4x + 1 = 0 to find the points on the graph where the tangent line has slope 1.