On what interval(s) is f(x)=x2-16/x-4 continuous? On what interval(s) is f(x)=x-

Question

On what interval(s) is f(x)=x2-16/x-4 continuous? On what interval(s) is f(x)=x-9/x2-7x-18 continuous? For the function f(x)=4x2+3x and the point [-3, 27], find the slope of the tangent line. 27 -14 8x-3 14 -21 What is the equation of the tangent line to f(x)=3/4x2 at the point (-2, 3)? y=3x+3 y=3/2x y=-3x+3 y=-3x-3 y=-3x-3 Find the derivative of f(x)=3x-6-6x-3 18x-7+18x-4 -18x-7+18x-4 -18x7-18x-4 -18x-5+18x-2 -18x-7+18x4 Find the derivative of y=7t8/5 21/5t4/5 56/32t3/5 56t7/5 56/5t3/5 56t2/5 Find the derivative of f(x)=5 5/2 10/ -5/ -5/2 5/ Find the derivative of f(x)=5/12x4 5/3x5 5/48x3 -20/3x5 20/3x5 -5/3x5

Explanation / Answer

(5)

we are given

f(x)=(x^2 -16)/(x-4)

firstly , we factor it

f(x)=((x-4)(x+4))/(x-4)

we can see that both are having common term =x-4

so, there will be hole at x=4

And we know any function can not be continuous at hole

Hence , this function is not continuous at x=4

and when denominator =0 ...we get asymptote

x+4=0

x=-4

this is also not contnuous at x=-4

so, function is continuous at all the points except at x=4 and x=-4

so, continuous function interval is

(-inf , -4 )U(-4, 4) U (4, inf)........Answer

(6)

f(x)=(x-9)/(x^2 -7x -18)

firstly , we factor it

f(x)=((x-9))/((x-9)(x+2))

we can see that both are having common term =x-9

so, there will be hole at x=9

And we know any function can not be continuous at hole

Hence , this function is not continuous at x=4

and when denominator =0 ...we get asymptote

x+2=0

x=-2

this is also not contnuous at x=-2

so, function is continuous at all the points except at x=9 and x=-2

so, continuous function interval is

(-inf , -2 )U(-2, 9) U (9, inf)........Answer

(7)

we are given

f(x)=4x^2 +3x

for finding slope , find derivative and then plug (-3, 27) or at x=-3

f'(x)=4*2x^(2-1) +3*1x^(1-1)

f'(x)=8x+3

now, we plug x=-3

slope=8*-3 +3

slope=-24+3

slope=-21.......Answer

(8)

f(x)=(3/4)x^2 at (-2,3)

firstly , find slope or m

for finding slope , find derivative and then we plug x=-2

f'(x)=(3/4)*2x^(2-1)

f'(x)=(3/4)*2x

slope=m=(3/4)*2*-2

m=-3

we have given point (-2, 3)

use point slope form of line

y-y1 = m(x-x1)

y-3=-3(x+2)

y-3 =-3x-6

equation of tangent line is :

y=-3x -3..........Answer

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