Question

f(x)=(x^7 - 7x^6)e^x. (a) f(x) is continuous on the interval: (b) f(x) is differentiable on the interval: (c) find all c on the interval 0,7 that satisfy the conclusion of the mean value theorem (mvt). (i) find the slope of the tangent line at c. f(c)=symbolic expression (ii) find the slope of the secant line. slope of secant line = symbolic expression (iii) if there are fewer than three cs that satisfy the conclusion of mvt on the interval, leave the remaining field(s) blank. c1 = c2 = c3 =

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 \(e^{c}
eq0\) 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) \(c_1 = 0\), \(c_2=\sqrt{42}\)