Help on a C3 Maths Question?

(a)assuming that e^(2x+1);

4e^(2x+1)=8

e^(2x+1)=2

ln(e^(2x+1))=ln2

2x+1=ln2

x={ln(2)-1}/2

x coordinate of P is {ln(2)-1}/2.

(b)y'=2*4*e^(2x+1)=8*e^(2x+1)

slope of tangent line at P;

=8*e^(2*(ln(2)-1)/2+1)

=8*e^(ln(2)-1+1)

=8*e^(ln2)=8*2=16

equation of tangent line;

(y-8)=16(x-(ln(2)-1)/2)

(y-8)=16x-8(ln(2)-1)

y=16x-8(ln(2)-2)

equation of tangent line;

y=16x-8(ln(2)-2)

(a) Since 8 is the y-coordinate of P, it goes into the left side of the equation. I'm also assuming (2x+1) is the exponent for e on the right side of the equation.

8 = 4e^2x+1

2^3 = 4e^2x+1

ln(2^3) =ln(4e^(2x+1))

3ln2 = ln(4) + ln(e^(2x+1))

3ln2 = ln(2^2) + 2x + 1

3ln2 = 2ln2 + 2x + 1

ln2 = 2x + 1

ln2 - 1 = 2x

(ln 2 - 1)/2 = x

(b) The slope of the curve is y' = 8e^(2x+1), and this is a.

a = 8e^(2(ln2 - 1)/2 + 1)

a = 8e^(ln2 - 1 + 1)

a = 8e^ln2

a = 8*2

a = 16

b is the y-intercept, the value of y when x = 0

y = 4e^(2x+1)

y =4e^(2(0)+1)

y = 4e

So the linear equation is y = 14x + 4e (or you can use a decimal approximation for the value of b)

Presumably thats e to the power (2x+1). Else I'm having trouble seeing the answer as a simple expression of ln2. But now the first bit is easy as y = 8 so e to the power 2x+1 is 2 and we just log both sides and then 2x+1 = ln2.

Second bit looks a bit of a mess still we have to differentiate to find the gradient of this line (thats the value of a).

y= e^w where w=2x+1

dy/dx = dy/dw . dw/dx = e^w.2

AND at this point e^w = 8 so the gradient dy/dx is 8x2 =16

So we have y = 16x + b and in principle we should be able to find b by substuting the values for x and y at the point we know. y is 8 of course and x is that expression in ln2 we found in the first part so it not pretty.

variety of f: because of the fact it rather is a parabola that opens upwards, there'll be a minimal element at (a million.5, 4.seventy 5) so the variety is y > or = 4.seventy 5 g(f(-a million)) first do f(-a million) that's (-a million)^2 - 3(-a million) + 7 = eleven then do g(eleven) = 2(eleven) -a million = 21 so g(f(-a million)) = 21 now you pick f(g(x)) to remedy fg(x) = 17 f(g(x)) = f(2x -a million) = (2x -a million)^2 -3(2x -a million) + 7 = 4x^2 -4x + a million - 6x + 3 + 7 = 4x^2 -10x + eleven now remedy 4x^2 -10x + eleven = 17 subtract 17 4x^2 -10x -6 = 0 element 2(x - 3)(2x + a million) = 0 chop up x - 3 = 0 --> x = 3 2x + a million = 0 --> x = -a million/2 {-a million/2,3}

(a) 8 = 4e^(2x + 1)

2 = e^(2x + 1)

ln 2 = 2x + 1

ln 2 - 1 = 2x

x = (1/2)(ln 2 - 1)

(b) dy/dx = 4e^(2x + 1) • 2 = 8e^(2x + 1) = 8e^(2a + 1)

b = 4e^(2a+ 1)

y - 4e^(2a + 1) = 8e^(2a + 1) (x - a)

y = 8e^(2a + 1) x - 8ae^(2a + 1) + 4e^(2a + 1)

y = 8e^(2a + 1) x + 4e^(2a + 1) (1 - 2a)