f(x)=(x^7 - 7x^6)e^x. (a) f(x) is continuous on the interval: (b) f(x) is differentiable on the interval…

Explanation:

Step1: Determine continuity

Polynomial - exponential product is continuous everywhere. So (f(x)=(x^{7}-7x^{6})e^{x}) is continuous on ((-\infty,\infty)).

Step2: Determine differentiability

The product of differentiable functions is differentiable everywhere. So (f(x)) is differentiable on ((-\infty,\infty)).

Step3: Find the derivative of (f(x))

Use the product - rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = x^{7}-7x^{6}) and (v = e^{x}). (u^\prime=7x^{6}-42x^{5}), (v^\prime = e^{x}), then (f^\prime(x)=(7x^{6}-42x^{5})e^{x}+(x^{7}-7x^{6})e^{x}=e^{x}(x^{7}- 42x^{5})). So (f^\prime(c)=e^{c}(c^{7}-42c^{5})).

Step4: Calculate the slope of the secant line

The slope of the secant line on ([a,b]=[0,7]) is (\frac{f(7)-f(0)}{7 - 0}). (f(0)=(0^{7}-7\times0^{6})e^{0}=0), (f(7)=(7^{7}-7\times7^{6})e^{7}=(7^{7}-7^{7})e^{7}=0). So the slope of the secant line is (\frac{0 - 0}{7-0}=0).

Step5: Solve for (c) using MVT

Set (f^\prime(c)=0), i.e., (e^{c}(c^{7}-42c^{5})=0). Since (e^{c}\neq0) for all real (c), we solve (c^{7}-42c^{5}=c^{5}(c^{2}-42)=0). So (c = 0), (c=\sqrt{42}), (c =-\sqrt{42}). But we are on the interval ([0,7]), so we discard (c=-\sqrt{42}). So (c_1 = 0), (c_2=\sqrt{42}).

Answer:

(a) ((-\infty,\infty)) (b) ((-\infty,\infty)) (c) (i) (f^\prime(c)=e^{c}(c^{7}-42c^{5})) (ii) (0) (iii) (c_1 = 0), (c_2=\sqrt{42})