Answer to Question #110585 in Algebra for atta

Question

functions f and g are defined by
f(x)=2x square +8x +1 for x is R
g(x)=2x-k for x is R
where k is constant
1)find the value of k for which the line y =g (x) is a tangent to the curve y=f(x).
2)in the case where k=-1 , find the set of values of x for which f(x)<g(x).
3)in the case where k=-1 , find g-1 f(x) and solve the equation g-1 f(x)=0.
4)express f(x) in the form 2(x+a)square + b ,where a and b are constant ,and hence state the least value of f(x)

Answer 1

solutions

given functions

f(x)=2x²+8x+1

g(x)=2x-k

1.) value of k for which the line y=g(x) is tangent to the curve y=f(x)

gradient function of the curve at a point is equal to the gradient of the tangent at that point.

hence:

f'( x)=4x+8

g'(x)=2

f'(x)=g'(x)

4x+8=2

x=-1.5

f(-1.5)=2(-1.5)²+8(-1.5)+1

f(-1.5)=-6.5

g(-1.5)=f(-1.5)

2(-1.5)-k=-6.5

-3-k=-6.5

dividing through by (-1) and make k the subject

k=3.5 or 7/2

2.) when k=-1 the functions become

f(x)=2x²+8x+1

g(x)=2x+1

values of x for f(x)<g(x)

2x²+8x+1<2x+1

putting like terms together implies

2x²+6x<0 implies 2x(x+3)<0

• hence x<0

x+3<0 implies x<-3

hence at x=[-2,-1] since at x= 0 and x=-3 f(x)=g(x)

hence x=[-2,-1] f(x)<g(x)

3.) where k=-1 find g^-1(f(x))

f(x)=2x²+8x+1

g(x)=2x+1

y=g(x)=2x+1

to find inverse substitute x with y and y with x then make y the subject

that is

x=2y+1

make y the subject you get y=(x-1)/2

then g^-1(x)=(x-1)/2

g^-1(f(x))=(2x²+8x+1-1)/2

g^-1(f(x))=x²+4x

hence solving g^-1(f(x))=0 we have

x²+4x=0 implies x(x+4)=0

• hence x=0
• x+4= 0 implies x=-4

4.) express f(x) in the form 2(x+a)²+ b

this means 2x²+8x+1=2(x+a)²+b

expanding 2(x+a)²+ b we get

2x²+8x+1=2(x²+2xa+a²)+b

open the brackets and equate the coefficients

2x²+8x+1=2x²+4xa+2a²+b

this implies 8x=4ax

• hence a=2

1=2a²+b but a=2 then 1=2(4)+b

• hence b=-7

therefore f(x)=2(x+2)²+(-7)