Find the instantaneous rate of change for the functions at x = -1

Question

a.) f(x)= 3x-2-4x²

b.) g(x) = 1/(x+2)

c.) h(x) = sqrt(x+5)

*sqrt = square root

Answer 1

a) f(x) = 3x -2 -4x^2

the instantaneous rate of change = the derivative of f(x)

take the derivative term by term and add them

derivative of 3x = 3,

derivative of a linear term is the coefficient

derivative of -2 = 0,

derivative of any constant term= 0

derivative of a non-linear term, f(x)=ax^b,

is f'(x) = abx^(b-1), bring down the exponent and make it into a coefficent,

then reduce the exponent by 1

with f(x)=-4x^2,

f'(x)=-4(2)x^(2-1) =-8x

f'(x) = 3-0-4(2)x

= 3-8x

replace x by -1

f'(-1) = 3-8(-1) = 3+8= 11

the instantaneous rate of change at a point = the slope of the tangent line at that point

f(x)= 3x -2 -4x^2 graphically is a downward opening parabola

you might try sketching the graph and find the point where x=-1 which is (-1,f(-1)) = (-1,-9)

It's on the left half of the parabola, where the slope gets relatively steep. slope =11 looks about right.

axis of symmetry is x=3/8, to the left of x=3/8, the slope is positive, such as at x=-1, the further left of the vertex, the steeper the slope of the tangent line to the parabola.

the vertex = max point = (3/8, f(3/8)) = (3/8, 9/8-2-9/16) = (3/8, -3/2)

c) h(x) = sqr(x+5)

the square root of a term is the same as the term raised to the 1/2 power

= (x+5)^(1/2)

take the derivative,

bring down the exponent and make it a coefficent,

reduce the exponent by 1

h'(x) = (1/2)(x+5)^(1/2-1)

= (1/2)(x+5)^(-1/2)

= (1/2)(1/(x+5)^(1/2)

=(1/2)(1/sqr(x+5)

= 1/2(sqr(x+5)

h'(-1) =1/2(sqr(-1+5)

=1/2(sqr4) = 1/2(2) = 1/4

this is another parabola, or rather the top half of a rightward opening parabola

with vertex = (-5, f(-5)) = (-5,0)

as x gets larger the slope of the tangent line gets smaller, and approaches zero with very large x

slope = 1/4 looks about right, if you look at a sketch of the graph

b) g(x) = 1/(x+2)

= (x+2)^-1

make the exponent into a coefficient,

and reduce the exponent by 1

g'(x) = -1(x+2)^(-1-1)

g'(x) = -1(x+2)^-2

= -1/(x+2)^2

g'(-1) = -1/(-1+2)^2

=-1/(1)^2 =-1/1 =-1

this function graphically is a rectangular hyperbola

with asymptotes at x=-2 and y=0

a rectangular hyperbola is the more standard vertical or horizontal hyperbola rotated by 45 degrees

at the point (-1,1) you're at the top right branch of the rectangular hyperbola, where all slopes are negative

(-1,1) is the vertex of that hyperbola branch, where the slope =-1