No one seems to wanna help me please?
1 answer:
For perpendicular questions, you need to find the reciprocal of the slope and change it's sign.
I sent a picture to show you what I mean
With your graph questions, remember
rYse over run
This is for your slope. for example, with number 19, go up 7 over the right 4(Because it's a positive number) starting from the y-intercept but also backtrack so it's a line and your y-intercept, b, the number after the slope is the point where the line meets the y- axis. So for number 19, you would plot a point at (0, -4)
I'm not entirely sure that's how the slope things go (The actual slope not the y-intercept) so I would suggest asking your teacher about that.
Sorry I couldn't help more! I just hope this makes you understand just a little bit more :)
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Answer:
a. 5 b. c. 148.5 d. 1/7
Step-by-step explanation:
Here is the complete question
Show all of your work, even though the question may not explicitly remind you to do so. Clearly label any functions, graphs, tables, or other objects that you use. Justifications require that you give mathematical reasons, and that you verify the needed conditions under which relevant theorems, properties, definitions, or tests are applied. Your work will be scored on the correctness and completeness of your methods as well as your answers. Answers without supporting work will usually not receive credit. Unless otherwise specified, answers (numeric or algebraic) need not be simplified. If your answer is given as a decimal approximation, it should be correct to three places after the decimal point. Unless otherwise specified, the domain of a function f is assumed to be the set of all real numbers for which f() is a real number Let f be an increasing function with f(0) = 2. The derivative of f is given by f'(x) = sin(πx) + x² +3. (a) Find f" (-2) (b) Write an equation for the line tangent to the graph of y = 1/f(x) at x = 0. (c) Let I be the function defined by g(x) = f (√(3x² + 4). Find g(2). (d) Let h be the inverse function of f. Find h' (2). Please respond on separate paper, following directions from your teacher.
Solution
a. f"(2)
f"(x) = df'(x)/dx = d(sin(πx) + x² +3)/dx = cos(πx) + 2x
f"(2) = cos(π × 2) + 2 × 2
f"(2) = cos(2π) + 4
f"(2) = 1 + 4
f"(2) = 5
b. Equation for the line tangent to the graph of y = 1/f(x) at x = 0
We first find f(x) by integrating f'(x)
f(x) = ∫f'(x)dx = ∫(sin(πx) + x² +3)dx = -cos(πx)/π + x³/3 +3x + C
f(0) = 2 so,
2 = -cos(π × 0)/π + 0³/3 +3 × 0 + C
2 = -cos(0)/π + 0 + 0 + C
2 = -1/π + C
C = 2 + 1/π
f(x) = -cos(πx)/π + x³/3 +3x + 2 + 1/π
f(x) = [1-cos(πx)]/π + x³/3 +3x + 2
y = 1/f(x) = 1/([1-cos(πx)]/π + x³/3 +3x + 2)
The tangent to y is thus dy/dx
dy/dx = d1/([1-cos(πx)]/π + x³/3 +3x + 2)/dx
dy/dx = -([1-cos(πx)]/π + x³/3 +3x + 2)⁻²(sin(πx) + x² +3)
at x = 0,
dy/dx = -([1-cos(π × 0)]/π + 0³/3 +3 × 0 + 2)⁻²(sin(π × 0) + 0² +3)
dy/dx = -([1-cos(0)]/π + 0 + 0 + 2)⁻²(sin(0) + 0 +3)
dy/dx = -([1 - 1]/π + 0 + 0 + 2)⁻²(0 + 0 +3)
dy/dx = -(0/π + 2)⁻²(3)
dy/dx = -(0 + 2)⁻²(3)
dy/dx = -(2)⁻²(3)
dy/dx = -3/4
At x = 0,
y = 1/([1-cos(π × 0)]/π + 0³/3 +3 × 0 + 2)
y = 1/([1-cos(0)]/π + 0 + 0 + 2)
y = 1/([1 - 1]/π + 2)
y = 1/(0/π + 2)
y = 1/(0 + 2)
y = 1/2
So, the equation of the tangent at (0, 1/2) is
c. If g(x) = f (√(3x² + 4). Find g'(2)
g(x) = f (√(3x² + 4) = [1-cos(π√(3x² + 4)]/π + √(3x² + 4)³/3 +3√(3x² + 4) + 2
g'(x) = [3xsinπ√(3x² + 4) + 18x(3x² + 4) + 9x]/√(3x² + 4)
g'(2) = [3(2)sinπ√(3(2)² + 4) + 18(2)(3(2)² + 4) + 9(2)]/√(3(2)² + 4)
g'(2) = [6sinπ√(12 + 4) + 36(12 + 4) + 18]/√12 + 4)
g'(2) = [6sinπ√(16) + 36(16) + 18]/√16)
g'(2) = [6sin4π + 576 + 18]/4)
g'(2) = [6 × 0 + 576 + 18]/4)
g'(2) = [0 + 576 + 18]/4)
g'(2) = 594/4
g'(2) = 148.5
d. If h be the inverse function of f. Find h' (2)
If h(x) = f⁻¹(x)
then h'(x) = 1/f'(x)
h'(x) = 1/(sin(πx) + x² +3)
h'(2) = 1/(sin(π2) + 2² +3)
h'(2) = 1/(sin(2π) + 4 +3)
h'(2) = 1/(0 + 4 +3)
h'(2) = 1/7