Calculus questions help please?

1) dy/dx at x = 4.... = 2 - 6(4) = -22

2) d f(x) / dx at x = 2.... = 6(2) + 2 = 14

3) In other words, what is the domain of y? x is all real numbers

4) f ' (x) = 0, = 3x^2 - 6x - 9 = 0, when x^2 - 2x - 3 = 0, (x-3)(x+1)=0, so x = 3 and x = -1 are the answers.

5) 2x + 4 = 10 when x = 6

6) dy / dx = 2x + 1. At x=3, dy / dx = 7 (= slope of the tangent)

1. To find a gradient you have to differentiate.

If y = 6 + 2x - 3x^2

then, y' = 2 - 6x (y' is the same as the gradient, m)

You then substitute the given x value into the differentiated function;

y' = m = 2 - 6(4)

= -22

2. Same as question 1;

y' = 6x + 2

y' = m = 6(2) + 2

= 14

3. Not sure what the question is asking. Are you sure you copied it out fully?

4. Differentiate the function;

f '(x) = 3x^2 - 6x - 9

You want to find the points where the gradient is zero so;

0 = 3x^2 - 6x - 9

Solve for x;

0 = (3x - 9)(x + 1)

x = 3 or -1

Substitute x values back into the ORIGINAL equation;

f(x) = (3)^3 - 3(3)^2 - 9(3) +1 or f(x) = (-1)^3 - 3(-1)^2 - 9(-1) +1

f(x) = -26 or 6

So the points are (3,-26) and (-1,6)

5. Same as above but with gradient of 10.

y' = 2x + 4

10 = 2x + 4

x = 3

y = (3)^2 + 4(3)

= 21

Answer is (3,21)

6. Differentiate the function;

y' = 2x + 1

Substitute the known x value into the equation;

y' = m = 2(3) + 1 = 7

The general equation for a straight line is y - y1 = m(x - x1), where (x1,y1) is a known point on the line;

In this case, (x1,y1) is (3,11) and m = 7.

Substitute these values into the general equation;

y - 11 = 7(x - 3)

y - 11 = 7x - 21

y = 7x - 10

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