mariahgriego4063 mariahgriego4063
  • 06-08-2017
  • Mathematics
contestada

Let $f(x) = x^2 + 4x - 31$. for what value of $a$ is there exactly one real value of $x$ such that $f(x) = a$?

Respuesta :

barnuts
barnuts barnuts
  • 18-08-2017

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

Answer Link

Otras preguntas

Given 4x2 – 7x = -3, what are the solutions??
Solve the inequality and then determine which graph is correct. 55 ≤ 24 + x
NEED HELP DUE IN 3 MINUTES!!!!
5. The watchman ........ go home now.6. .......I borrow your book for a week?7. The lawyer will ........meet his clients in the evening.8. The labourers .......
An investor purchased on margin Orange Computer for $30 a share. The stock's price subsequently increased to $47 a share at which time the investor sold the sto
How to calculate raw score from standard score and percentile rank?
Organizations can use common core security principles recommended as industry best practices when developing policies, standards, baselines, procedures, and gui
2(x+3)=x-4 please help me <3
how much can your bladder hold before it pops
Find an equation of a line that is parallel to y = 5x - 21 and passes through the point (3, 7)