Math  /  Calculus

QuestionSuppose f(x)=x3+4,x[0,1]f(x)=x^{3}+4, x \in[0,1]. (a) Find the slope of the secant line connecting the points (x,y)=(0,4)(x, y)=(0,4) and (1,5)(1,5). (b) Find a number c(0,1)c \in(0,1) such that f(c)f^{\prime}(c) is equal to the slope of the secant line you computed in (a), and explain why such a number must exist in ( 0,1 ).

Studdy Solution
Use the Mean Value Theorem, which states that if f(x) f(x) is continuous on [a,b][a, b] and differentiable on (a,b)(a, b), then there exists at least one c(a,b) c \in (a, b) such that: f(c)=f(b)f(a)baf'(c) = \frac{f(b) - f(a)}{b - a}
In this case, a=0 a = 0 and b=1 b = 1 , and we have already found that: f(1)f(0)10=1\frac{f(1) - f(0)}{1 - 0} = 1
We need to find c c such that: f(c)=13c2=1f'(c) = 1 \quad \Rightarrow \quad 3c^2 = 1
Solve for c c : c2=13c=13c^2 = \frac{1}{3} \quad \Rightarrow \quad c = \sqrt{\frac{1}{3}}
Since c c must be in the interval (0,1) (0, 1) , and 130.577\sqrt{\frac{1}{3}} \approx 0.577, c c is indeed in (0,1) (0, 1) .
The number c c that satisfies the condition is:
13 \boxed{\sqrt{\frac{1}{3}}}

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