a)
(4−8x)2=(4)2−2×4×8x+(8x)2=16−64x+64x2
This confirms the given expression.
b) To find the value of ∫30(4−8x)2dx, we use the result from part (a), where:
(4−8x)2=16−64x+64x2
Thus, the integral becomes:
∫30(16−64x+64x2)dx
∫30(16−64x+64x2)=16x−32x2+643x3dx
(16−32(3)2+64333)−16=336
a) To find the equation of the tangent to f(x) at point Q(2,14), we first need to determine the gradient of the function at x=2.
f′(2)=9(2)2−4(2)+1
f′(2)=36−8+1=29
So, the gradient of the tangent at x=2 is 29.
Using the point-slope form of the equation of the tangent line:
y−14=29(x−2)
y−14=29x−58
y=29x−44
Thus, the equation of the tangent line at point Q is y=29x−44.
b) To find the equation of f(x), we integrate f′(x).
f(x)=3x3−2x2+x+C
To find the constant C, use the point Q(2,14):
14=3(2)3−2(2)2+2+C
C=14−18=−4
So, the equation of f(x) is:
f(x)=3x3−2x2+x−4
a)
a=∫32(2x2−3x−2)dx
b)
∫(2x2−3x−2)dx=2x33−3x22−2x
Evaluating this from x=2 to x=3:
Substitute x=3:
2(3)33−3(3)22−2(3)=543−272−6
=18−13.5−6=−1.5
Substitute x=2:
2(2)33−3(2)22−2(2)=163−122−4
=163−6−4=163−10=−143
The area a is:
a=(−1.5)−(−143)=−1.5+143=196
So the area a is:
a=196
c) To find the value of c, which is measured as c=b2a, we first need to find b. To do that, we need to find the top side of the rectangle which it forms, which can be done by finding f(3)
f(3)=7∗1=7
So the graph of this function looks as follows:
So, we have that:
b=7−196=236
So, finally we have that
c=b2a=236196=2319
Closea) To find the equation of f(x), we integrate the given gradient function f′(x)=kx+5:
f′(x)=kx+5
Integrating f′(x) with respect to x:
f(x)=∫(kx+5)dx=kx22+5x+C
Given that f(x) cuts the y-axis at y=2, we know f(0)=2:
f(0)=k⋅022+5⋅0+C=2⟹C=2
So, the equation becomes:
f(x)=kx22+5x+2
Given that f(x) cuts the x-axis at x=2, we know f(2)=0:
f(2)=k⋅222+5⋅2+2=0⟹2k+10+2=0⟹2k+12=0⟹k=−6
Thus, the function f(x) is:
f(x)=−6x22+5x+2=−3x2+5x+2
b) This can be easily found in the GDC:
Area=∫20(−3x2+5x+2)dx=6
c) To find the tangent at x=1, we need the slope at x=1. The slope is given by f′(x):
f′(x)=kx+5=−6x+5
Evaluating the slope at x=1:
f′(1)=−6⋅1+5=−1
The slope of the tangent line at x=1 is −1.
Next, we find the y-coordinate at x=1:
f(1)=−3(1)2+5⋅1+2=−3+5+2=4
So the point of tangency is (1,4).
Using the point-slope form of the equation of a line:
y−y1=m(x−x1)
Where m is the slope and (x1,y1) is the point of tangency:
y−4=−1(x−1)
Simplifying:
y−4=−x+1⟹y=−x+5
Closea) To find the equation of the tangent to f(x)=−2x3+6x2 at x=3 we have to first find the derivative:
f(x)=−2x3+6x2
f′(x)=ddx(−2x3+6x2)=−6x2+12x
f′(3)=−6(3)2+12(3)=−54+36=−18
So, the slope of the tangent at x=3 is −18.
Now, we find the y-coordinate at x=3:
f(3)=−2(3)3+6(3)2=−54+54=0
So, the point of tangency is (3,0).
Then, we find the tangent line:
y−y1=m(x−x1)
Substituting m=−18, x1=3, and y1=0:
y−0=−18(x−3)⟹y=−18x+54
b) To find the shaded area we first have to find the x-intercepts:
f(x)=−2x3+6x2=x2(−2x+6)
x2(−2x+6)=0
x=0,x=3
Now, we can plug the function into the GDC and find the area:
So, the area is equal to 13.5.
c)
Area of triangle=12×base×height
Given the vertices A(1,0), B(5,0), and C(3,y), the base AB is 5−1=4.
12×4×y=13.5
2y=13.5⟹y=13.52=6.75
The y-coordinate of point C is 6.75.
Closea)
4√z+20√z=4√z√z+20√z
We know that 4√z√z=4 and 20√z=20z−12. Therefore, the expression becomes:
4+20z−12
This matches the form 4+20zp where p=−12.
b) To find the value of ∫1614√z+20√zdz, we use the simplified form 4+20z−12:
∫161(4+20z−12)dz
∫161(4+20z−12)dz=∫1614dz+∫16120z−12dz
Evaluating the integrals:
∫1614dz=4z|161=4(16)−4(1)=64−4=60
∫16120z−12dz=20(2z12)|161=40z12|161
Evaluating this at the bounds:
40z12|161=40(1612)−40(112)=40(4)−40(1)=160−40=120
Adding the results of both integrals:
60+120=180
Thus, the value of the integral is 180.
Closea) As it can be seen, we have −x2 for f(x), meaning that i's concave down, so it will be green. For g(x) we have x2, so it is concave up, which means that it is the red curve.
b) To find the points of intersection of the functions, we set f(x)=g(x):
−x2+3x−11=x2−5x−5
−x2+3x−11−x2+5x+5=0
−2x2+8x−6=0
x2−4x+3=0
(x−1)(x−3)=0
So, the solutions are:
x=1andx=3
Since a>b:
a=3andb=1
c) To find the area enclosed by the two functions between x=1 and x=3, we integrate the difference f(x)−g(x) from 1 to 3. We start with f(x) as this is the function on top:
Area=∫31[f(x)−g(x)]dx
Substitute f(x) and g(x):
f(x)=−x2+3x−11
g(x)=x2−5x−5
f(x)−g(x)=(−x2+3x−11)−(x2−5x−5)
f(x)−g(x)=−2x2+8x−6
Now, integrate −2x2+8x−6 from 1 to 3:
∫31(−2x2+8x−6)dx
∫(−2x2+8x−6)dx=−2x33+4x2−6x
Evaluate this from 1 to 3:
[−2x33+4x2−6x]31
First, evaluate at x=3:
−2(3)33+4(3)2−6(3)=−543+36−18=−18+36−18=0
Next, evaluate at x=1:
−2(1)33+4(1)2−6(1)=−23+4−6=−23−2=−2+63=−83
The area is:
0−(−83)=83
Thus, the area enclosed by the two functions is 83 square units.
CloseTo solve the integral, we use the trigonometric identity:
cos2θ=1+cos(2θ)2
Applying this identity with θ=x4:
cos2(x4)=1+cos(x2)2
Substitute this into the integral:
∫2π0cos2(x4)dx=∫2π01+cos(x2)2dx
Split this into two separate integrals:
∫2π01+cos(x2)2dx=12∫2π01dx+12∫2π0cos(x2)dx
Evaluate each integral separately:
1. Integral of 1:
12∫2π01dx=12[x]2π0=12⋅(2π−0)=π
2. Integral of cos(x2):
Use substitution u=x2, hence du=12dx or dx=2du. Adjust the limits:
When x=0, u=0. When x=2π, u=π.
12∫2π0cos(x2)dx=12∫π0cos(u)⋅2du=∫π0cos(u)du
Integrate:
∫π0cos(u)du=sin(u)|π0=sin(π)−sin(0)=0−0=0
Combine the results:
12∫2π01dx+12∫2π0cos(x2)dx=π+0=π
Thus, the value of the integral is:
∫2π0cos2(x4)dx=π
CloseTo find the function f(x), we need to integrate f′(x):
f(x)=∫5x7x2+3dx
Substitute u=7x2+3:
du=14xdx or dx=du14x
Substitute into the integral:
∫5x7x2+3dx=∫5xu×du14x
=514∫duu
=514ln|u|+C
Substitute back u=7x2+3:
f(x)=514ln|7x2+3|+C
To find C, use the point Q(1,3):
3=514ln|7×12+3|+C
3=514ln|10|+C
Solving for C:
C=3−514ln10
The final equation for f(x) is:
f(x)=514ln|7x2+310|+3
Closea)
f(x)=−4cos2(x)−2sin2(x)+2
Rewrite f(x) using the identity sin2(x)=1−cos2(x):
f(x)=−4cos2(x)−2(1−cos2(x))+2
f(x)=−4cos2(x)−2+2cos2(x)+2
f(x)=−2cos2(x)
Set f(x)=0:
−2cos2(x)=0
cos2(x)=0
cos(x)=0
The cosine function is zero at x=π2 within the interval 0≤x≤π. Therefore, the root is:
x=π2
b)
f(x)=−2cos2(x)
Using the chain rule:
ddx[cos2(x)]=2cos(x)(−sin(x))=−2cos(x)sin(x)
Thus:
f′(x)=−2×(−2cos(x)sin(x))=4cos(x)sin(x)
f′(x)=2sin(2x)
So, a=2 and b=2.
c)
2sin(2x)=0
sin(2x)=0
The general solution for sin(2x)=0 is:
2x=nπ for integer n
x=nπ2
Within the interval 0≤x≤π, the values are x=0 and x=π2.
We substitute these x-values into f(x) to find the corresponding y-values:
For x=0:
f(0)=−2cos2(0)=−2⋅12=−2
For x=π2:
f(π2)=−2cos2(π2)=−2⋅02=0
So, the coordinates are: (0,−2), (π2,0)
d) We use what we found previously:
−4cos2(x)−2sin2(x)+2=−2cos2(x)
Rewrite cos2(x) using cos2(x)=1+cos(2x)2:
∫π0−2(1+cos(2x)2)dx=∫π0−1−cos(2x)dx
Integrate term by term:
∫π0−1dx−∫π0cos(2x)dx
Using the substitution u=2x, du=2dx, so dx=du2:
∫π0cos(2x)dx=12∫2π0cos(u)du
∫cos(u)du=sin(u)
12[sin(u)]2π0=12[sin(2π)−sin(0)]=0
So:
∫π0−1dx=−x|π0=−π
∫π0cos(2x)dx=0
Thus:
∫π0−2cos2(x)dx=−π
Closea) To find the inverse function g−1(x), we start with the function g(x)=√3x+4. Set y=√3x+4 and solve for x:
y=√3x+4
Square both sides to eliminate the square root:
y2=3x+4
Rearrange to solve for x:
3x=y2−4
x=y2−43
Switch x and y to find the inverse function:
g−1(x)=x2−43
b) To find the point of intersection we can plug both functions into the GDC:
The coordinates of intersection are (4,4).
c)
g′(x)=12(3x+4)−1/2×3=32√3x+4
d) We first find the derivative of g−1(x):
g−1(x)=x2−43
g−1(x)′=2x3
Equalize both derivatives:
32√3x+4=2x3
Now we can plug them both into the GDC as well:
So, the gradients are the same for x=0.874
e) The area enclosed will look like the one presented on the graph below:
To find this area, we calculate the following integral. This can be easily done in the GDC:
∫40(√3x+4−x2−43)dx