calculus and curve sketching q's (1 Viewer)

[Kaatie]

1. Determine the nature of the stationary pt of the curve y=(9-x^2)^1/2 is meant to be max at x=0

2. Determine the range of y = f(x), -3<x< 3 for the function f(x)= x^4-4x+7 i keep getting 76<x<100 but the answer is 4<x<100

3. Find the equation of the tangent to the curve y=1/x at the point where x=2, the answer is x+4y-4=0, again i keep getting the wrong answer

4. At what pt on the curve y=2x^2-7x+5 is the tangent parallel to the line y=5x-2? have not been taught this yet.

5. Find the equation of the normal to the curve y=1/(2x-1) at (2,1/3), again keep getting wrong answer is meant to be 4x +y-18=0

6. For what values of x is the curve y=x^3+9x-3 increasing
y'=3x^2+9
inc when y'>0
3x^2+9>0
x^2>-3
now what?? is meant to be all values of x


i know it is alot of questions so i do not expect all of them done but if anyone could have a go at one it would be great

 

[bored of sc]

3. Find the equation of the tangent to the curve y=1/x at the point where x=2, the answer is x+4y-4=0, again i keep getting the wrong answer

y = x^-1
y' = -x^-2
when x=2
y' = -(2)^-2
= -1/2²
= -1/4

when x=2
y = 1/2

thus, m = -1/4 at the point (2,1/2)
Now use the one point formula:
y-1/2 = -1/4(x-2)
4y-2 = -x+2
x+4y-4 = 0

 

[bored of sc]

4. At what pt on the curve y=2x^2-7x+5 is the tangent parallel to the line y=5x-2? have not been taught this yet.

4.
y = 2x²-7x+5
y' = 4x-7

Now for this tangent to be parallel to the line y = 5x-2 they must each have the same gradient i.e. a gradient of 5.

So 4x-7 = 5
4x = 12
x = 3

when x = 3
y = 2*9-7*3+5
= 2

therefore the point on the curve is (3,2)

 

[bored of sc]

5. Find the equation of the normal to the curve y=1/(2x-1) at (2,1/3), again keep getting wrong answer is meant to be 4x +y-18=0

y = (2x-1)^-1
y' = -(2x-1)^-2*2
= -2/(2x-1)²

since we are given the point (2,1/3) sub in x=2 into the derivative to find the gradient of the TANGENT

y' = -2/3²
= -2/9

now the normal is perpendicular to the tangent i.e. m(tangent)*m(normal) = -1

so -2/9*m = -1
m = 9/2 = 4.5

So the point is (2,1/3) and the gradient of the normal at this point is 9/2

Now use the one point formula to produce the equation:
y-1/3 = 9/2(x-2)
6y-2 = 27x-54
27x-6y-52 = 0