1) The graph of the function f(x) = 4x^3 - 48x has one local maximum point and o

Question

1) The graph of the function f(x) = 4x^3 - 48x has one local maximum point and one local minimum point. Find these points using the derivative test.

2) The graph of the following function has one relative maximum point and one relative minimum point. Find these points using the first derivative test. f(x) = x^3 + 9x^2 + 15x + 1

3) The graph of the function has one relative extreme point. Plot this point and check the concavity there. Using only this information, sketch the graph. f(x) = 3x^2 - 7

4) The graph of the function has one relative maximum and one relative minimum point. Plot these two points and check the concavity there. Using only this information, sketch the graph. f(x) = x^3 + 9x^2 + 15x

Explanation / Answer

1) given function f(x) = 4x^3 - 48x

differentiate with respect to x

f '(x)=12x2-48

for critical point f '(x)=0

12x2-48=0

=>x=-2, x =2

for x<-2, f '(x)>0

for -2<x<2, f '(x)<0

for x>2, f '(x)>0

so function has  local maximum point at x =-2, local minimum point at x=2

f(-2)=64,f(2)=-64

local maximum point is (-2,64), local minimum point is (2,-64)

2)given function f(x) = x^3 + 9x^2 + 15x + 1

differentiate with respect to x

f '(x)=3x2 + 18x + 15

for critical point f '(x)=0

3x2 + 18x + 15 =0

3(x2 + 6x + 5)=0

3(x+5)(x+1)=0

=>x=-5, x =-1

for x<-5, f '(x)>0

for -5<x<-1, f '(x)<0

for x>-1, f '(x)>0

so function has   relative maximum point at x =-5,  relative minimum point at x=-1

f(-5)=26,f(-1)=-6

relative  maximum point is (-5,26), relative  minimum point is (-1,-6)

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.