SOLUTION: For the following function {{{g(x)=-4x^2+8x+3}}} I am to determine wherether this function has a minimum or maximum value and I am to find the value. I have four problems like this

Algebra ->  Functions -> SOLUTION: For the following function {{{g(x)=-4x^2+8x+3}}} I am to determine wherether this function has a minimum or maximum value and I am to find the value. I have four problems like this      Log On


   

Question 112446: For the following function g%28x%29=-4x%5E2%2B8x%2B3 I am to determine wherether this function has a minimum or maximum value and I am to find the value. I have four problems like this and this is the first one. I am having a hard time getting started. Can you please help me with this problem so that i can learn how to compelete the rest. Thank you so much for all your help!!!!
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
The way you go about this depends on whether you are studying calculus and know how to take the first and second derivatives of a polynomial function or not. Not knowing, I'll do it two different ways:

First the calculus way.
The first derivative gives us the slope of a line tangent to the curve at any point in the domain of the function. Since the tangent line at a local minimum or maximum on the curve is horizontal, and the slope of a horizontal line is zero, you simply set the first derivative equal to zero and solve to find the x-coordinate of the extrema. In this case, given:

g%28x%29=-4x%5E2%2B8x%2B3
dg%28x%29%2Fdx=-8x%2B8
-8x%2B8=0
-8x=-8
x=1

Knowing the x-coordinate of the point allows us to find the y-coordinate by evaluating g(1).

g%281%29=-4%281%29%5E2%2B8%281%29%2B3=-4%2B8%2B3=7

And the critical point is the point (1, 7)

Now we still have to determine whether the extreme point is a minimum or a maximum. If it is a maximum, the slope of a tangent to the curve will change from positive to negative as x moves left to right across the critical point. Conversely, if it is a minimum, the tangent slope will change from negative to positive as you move across. The second derivative, being the rate of change of the rate of change, evaluated at the x-coordinate of the critical point will tell us by its sign which way it is going. If the second derivative at the point is negative, then we are at a maximum. If the second derivative is positive, it is a minimum.

g%28x%29=-4x%5E2%2B8x%2B3
dg%28x%29%2Fdx=-8x%2B8
d%5E2g%28x%29%2Fdx%5E2=-8

Since the second derivative is negative, this is a local maximum.


Now for the non-calculus way. Since we are dealing with a 2nd degree polynomial, we know that if we graphed it, we would see a parabola. The vertex (h, k) of a parabola f%28x%29=ax%5E2%2Bbx%2Bc is determined by the following:

h=%28-b%29%2F2a, and
k=f%28h%29=ah%5E2%2Bbh%2Bc

In our case:
h=%28-8%29%2F2%28-4%29=1, and
k=f%281%29=-4%281%29%5E2%2B8%281%29%2B3=7

So the vertex, which is the critical point, is at (1, 7)

Since the coefficient on the x%5E2 term is negative, i.e. -4, we know that the parabola is concave down. That means the vertex is at the top and it represents a maximum.

Let's see if all this makes sense graphically:
graph%28400%2C400%2C-3%2C3%2C-2%2C8%2C-4x%5E2%2B8x%2B3%2C7%29

Looks good to me. How about you?


Hope this helps,
John