Let $f(x) = x^2 + 4x - 31$. for what value of $a$ is there exactly one real value of $x$ such that $f(x) = a$?
1 answer:
To solve for this, we need to find for the value of x
when the 1st derivative of the equation is equal to zero (or at the
extrema point).
So what we have to do first is to derive the given
equation:
f (x) = x^2 + 4 x – 31
Taking the first derivative f’ (x):
f’ (x) = 2 x + 4
Setting f’ (x) = 0 and find for x:
2 x + 4 = 0
x = - 2
Therefore the value of a is:
a = f (-2)
a = (-2)^2 + 4 (-2) – 31
a = 4 – 8 – 31
a = - 35
You might be interested in
For all the dogs: you have 2.65+2.46+3.67+2.91 = 11.69%
Now, the probability of getting a chihuahua would be 3.44/11.69=0.2942= 24.92
Step-by-step explanation:
there ya go!!
Answer:
C: 51%
Step-by-step explanation:
There is a 60% that the weather is good and 85% of going to the concert so you multiply the two because of probability. 0.6 x 0.85 = 0.51
Answer:
1 2 3 6 7 14 21 42Step-by-step explanation:
all of them are divisible by 42
K-12=28...? I think this is what your question is asking for
Answer:
(1,-1)
Step-by-step explanation:
X goes up +1 and Y goes down -1
