So all you need to do if find the min of the function:
f(x)=0.04x^3-4x^2-176x
f'(x)=0.12x^2-8x-176
The only zero for this is x=84.105 <---first
Plug this into f(x) to get f(84.105)=-19300 <---second
The answer would be:
y = 5
x = 4
Answer:
i did the first
Step-by-step explanation:
1st way
Standard form: a(X-h)²+k = ( -2/3X² -16/3X -32/3) +32/3 -17/3 = -2/3(X +4)² +5
y = -2/3*X^2-16/3*X-17/3
X = -4 ±√( 15/2) = -6.7386, or -1.2614
Axis of symmetry: X= -4; Vertex (maximum)=(h,k)=( -4, 5); y-intercept is (0,-5.66666666667)
two real roots: X=-1.2613872124776866 and -6.738612787477313
Answer:
f(g(9)) = 945/16
Step-by-step explanation:
To find f(g(x)), you have to substitute g(x) wherever there is an x in f(x).
g(x) = x + 3/4
f(x) = x² - 4x - 3
f(g(x)) = (x + 3/4)² - 4(x + 3/4) - 3
f(g(x)) = x² + 3/2x + 9/16 - 4x + 3 - 3
f(g(x)) = x² - 5/2x + 9/16 + 3 - 3
f(g(x)) = x² - 5/2x + 9/16
Now, put a 9 wherever there is an x in f(g(x)).
f(g(9)) = (9)² - 5/2(9) + 9/16
f(g(9)) = 81 - 5/2(9) + 9/16
f(g(9)) = 81 - 45/2 + 9/16
f(g(9)) = 117/2 + 9/16
f(g(9)) = 945/16
The mode is 3. Mode- the number that appears the most. 3 appears the most out of the set of numbers
