Question

question 1
consider the function ( f(x)=x^{2}-3x + 5 ). what is a value of ( -1lt clt3 ) such that
( \frac{f(3)-f(-1)}{3-(-1)}=f^{prime}(c) )?
enter your answer below (write dne if there is no value of ( c )).

Explanation:

Step1: Calculate \(f(3)\) and \(f(-1)\)

Substitute \(x = 3\) into \(f(x)=x^{2}-3x + 5\):
\(f(3)=3^{2}-3\times3 + 5=9 - 9+5 = 5\)
Substitute \(x=-1\) into \(f(x)=x^{2}-3x + 5\):
\(f(-1)=(-1)^{2}-3\times(-1)+5=1 + 3+5=9\)
Then \(\frac{f(3)-f(-1)}{3-(-1)}=\frac{5 - 9}{4}=\frac{-4}{4}=-1\)

Step2: Find the derivative \(f^{\prime}(x)\)

Using the power rule \((x^{n})^\prime=nx^{n - 1}\), for \(f(x)=x^{2}-3x + 5\), \(f^{\prime}(x)=2x-3\)

Step3: Solve for \(c\)

Set \(f^{\prime}(c)=-1\), so \(2c-3=-1\)
Add \(3\) to both sides: \(2c=-1 + 3=2\)
Divide both sides by \(2\): \(c = 1\)

Answer:

\(1\)