example 5 suppose that ( f(0)=-8 ) and ( f^{prime}(x) leq 9 ) for all values of ( x ). how large can ( f(3)…

example 5 suppose that ( f(0)=-8 ) and ( f^{prime}(x) leq 9 ) for all values of ( x ). how large can ( f(3) ) possibly be? solution we are given that ( f ) is differentiable (and therefore continuous) everywhere. in particular, we can apply the mean value theorem on the interval ( 0,3 ). there exists a number ( c ) such that ( f(3)-f(0)=f^{prime}(c)(quad-0) ) so ( f(3)=f(0)+f^{prime}(c)=-8+f^{prime}(c) ). we are given that ( f^{prime}(x) leq 9 ) for all ( x ), so in particular we know that ( f^{prime}(c) leq ). multiplying both sides of this inequality by 3, we have ( 3 f^{prime}(c) leq ), so ( f(3)=-8+quad f^{prime}(c) leq-8+ ) the largest possible value for ( f(3) ) is

example 5 suppose that ( f(0)=-8 ) and ( f^{prime}(x) leq 9 ) for all values of ( x ). how large can ( f(3) ) possibly be? solution we are given that ( f ) is differentiable (and therefore continuous) everywhere. in particular, we can apply the mean value theorem on the interval ( 0,3 ). there exists a number ( c ) such that ( f(3)-f(0)=f^{prime}(c)(quad-0) ) so ( f(3)=f(0)+f^{prime}(c)=-8+f^{prime}(c) ). we are given that ( f^{prime}(x) leq 9 ) for all ( x ), so in particular we know that ( f^{prime}(c) leq ). multiplying both sides of this inequality by 3, we have ( 3 f^{prime}(c) leq ), so ( f(3)=-8+quad f^{prime}(c) leq-8+ ) the largest possible value for ( f(3) ) is

Answer

Explanation:

Step1: Apply the Mean Value Theorem

By the Mean Value Theorem, (f(3)-f(0)=f^{\prime}(c)(3 - 0)) for some (c\in[0,3]). Given (f(0)=-8), we can rewrite it as (f(3)=-8 + 3f^{\prime}(c)).

Step2: Use the inequality for (f^{\prime}(x))

We know that (f^{\prime}(x)\leq9) for all (x). Substituting (x = c) (since (c\in[0,3])), we get (f^{\prime}(c)\leq9).

Step3: Find the upper - bound for (f(3))

Substitute (f^{\prime}(c)\leq9) into the equation (f(3)=-8 + 3f^{\prime}(c)). Then (f(3)\leq-8+3\times9). Calculate ( - 8+3\times9=-8 + 27).

Answer:

(19)