If a continuous, differentiable function f is equal to 2 at x = 3 and at x = 5, what can you say about f ' on [3, 5]?

Consider the given points of function$\left[3,5\right]$.

Function $f\left(x\right)=x^2-8x+15=\left(x-3\right)\left(x-5\right)$has roots at $x=3,x=5$

Since f is a polynomial, it is continuous and differentiable. In particular, it is continuous on [3, 5] and differentiable on (3, 5). Therefore Rolle’s Theorem applies to the function f ,and we can conclude that there must exist some value of c ∈ (3, 5) for which f '(c) = 0. At this value of c the graph of f will have a horizontal tangent line. Rolle’s Theorem tells us that there exists some c ∈ (3, 5) where f '(c) =0, but it doesn’t tell us exactly where. We can find such a c by solving the equation f '(x) = 0. Since $f\left(x\right)=x^2-8x+15$, we have $f\text{'}\left(x\right)=2x-8$, which is equal to zero when x =4 Therefore f has a horizontal tangent line at c =4 which is in the interval (3, 5). The following function illustrates that $f\left(x\right)=x^2-8x+15$does appear to have a horizontal tangent line at x =4

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The function is $f\left(x\right)=x^2-8x+15$to find the point c differentiate the given function and put equal to zero .

$$\begin{gathered} f\left(x\right)=x^2-8x+15 \\ f\text{'}\left(x\right)=2x-8 \end{gathered}$$

Further simplify .

$f\text{'}\left(x\right)=0$

$$\begin{gathered} 2x-8=0 \\ 2x=8 \\ x=4 \end{gathered}$$

Therefore the value of x lie between the points$\left[3,5\right]$.