# Pre-Calculus

posted by .

Find the maximum value, and where it occurs for eAch function.
G(x)= (square root -x^2+4x+12)

• Pre-Calculus -

the vertex of the parabola
y = -x^2 + x + 12

x^2 - x = -y + 12

x^2 - x + 1/4 = -y+ 12.25

(x-1/2)^2 = -1 (y-12.25)

vertex at (1/2 , 12.25)

there -x^2 + 4 x + 12.25 = -1/4+2+12.25
= 14
G = sqrt(14)

• Pre-Calculus -

4x sort (25−x2)

• Pre-Calculus -

nvm that is not correct! i was trying to ask a question

## Similar Questions

1. ### Pre Calculus

Find the roots of the function g(x)=x^3 +16x a)0,4i, -4i b)0,-4,-4i c)0,4,4i d)0,4,-4 The equation can be rewritten x(x^2 + 16) = 0 The function is zero if x = 0, OR if x^2 = -16 That should give you a hint of what the roots are. So …
2. ### calculus

I have three questions I'm having a terrible time with: 1)Find, if possible, the absolute maximum value and where it occurs for f(x)=ln(xe^-x) on (0,infinity). 2)Find the value(s) of "c" guaranteed by the Mean Value Theorem for the …
3. ### calculus

Suppose that f(x)= square root of x and g(x)= 3x-4 and you are asked to evaluvate the composite function f(g(0)). What is the complication that occurs in trying to find that composite function value?
4. ### Pre calculus

find the domain of the given function f(x)=square root 9-x
5. ### pre-calculus

Please diregard the other one the square root sign did not come out right. find the complete function f o g f(x)=x^2+1;g(x)= square root x-4
6. ### college pre-calculus

simplify: square root (4/3)-square root (3/4) A. 2-sqaure root(3)/2 square root (3) B. square root (7/12) C. square root (3)/6 D. 2/square root (3) which answer?
7. ### Pre-Calculus

Solve 0 = 3x^2 + 5x -1 by completing the square. Express your answer as exact roots. 3(x^2 + 5/3 + 25/36) = -1 (x+ 5/6)^2 = 37/36 sq root( x + 5/6)^2 = sq root (37/36) x + 5/6 = +- sq root 37/6 x = -5/6 +- square root 37 /6 <--- …
8. ### Calculus

A rectangle is bounded by the x axis and the semicircle = square root 25-x^2. What length and width should the rectangle have so that its area is a maximum?
9. ### calculus

Find the absolute extrema of the function (if any exist) on each interval. (If an answer does not exist, enter DNE.) f(x) = square root of (25 − x^2) (a) [−5, 5] minimum (x, y) = (smaller x-value) (x, y) = (larger x-value) …
10. ### calculus

Is this the correct answers for these questions Verify the means value theorem holds on the interval shown. Then, find the value c such that f'(c)=(f(b)-f(a))/(b-a) b.f(x)=x^3=x-4 on [-2,3] c= square root 7/3 c. f(x)= x^3 on [-1,2] …

More Similar Questions