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h(x)= definite integral of ln(t^2)dt between 1 and x. a) Evaluate h(1), h'(1), and h"(1). b) Does h have a relative minimum, maximum, or neither at x = 1? Justify your answer. c) Let k = h-1. Evaluate k'(0).
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\[h(x)=\int\limits_1^x \ln(t^2) dt \] \[h(1)=\int\limits_1^1\ln(t^2) dt=?\]
By the fundamental theorem of calculus, \[\frac{d}{dx}\int_c^{g(x)}f(t)~dt=f(g(x))\cdot g'(x)\] So, \[h(x)=\int_1^x\ln t^2~dt~~\Rightarrow~~\begin{cases}h'(x)=\ln x^2\\h''(x)=\dfrac{2}{x}\end{cases}\] Part (b) is an application of the first derivative test. For part (c), you have \(k=h-1\), which gives you \[\begin{align*}k(x)&=(h-1)(x)\\ &=h(x)-1\\ &=\int_1^x\ln t^2~dt-1\\ k'(x)&=\ln x^2 \end{align*}\]
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